Inaugural Dissertation

INAUGURAL DISSERTATION

zur Erlangung der Doktorwürde der Naturwissenschaftlich-Mathematischen Gesamtfakultät der Ruprecht-Karls-Universität Heidelberg

vorgelegt von M.Sc. Yamidt Bermúdez Tobón aus Montebello (Ant.), Kolumbien Tag der mündlichen Prüfung:

Thema An eﬃcient algorithm to compute an elliptic curve from a corresponding function ﬁeld automorphic form

Gutachter: Prof. Dr. Gebhard B¨ockle

Acknowledgements

I thank my supervisor Prof. Dr. Gebhard B¨ockle for his guidance and support. Also I would like to thank to Dr. Juan Cervi˜no, he provided sound support and was always willing to help whenever I asked, during the whole core of this work. We had a lot of interesting and enthusiastic discussions which often contained useful ideas. Without their help this thesis would not have been possible. During this work I was supported by the DAAD for three and one half years and the DFG project Nr D.110200/10.005 from 01.01.2013 to 30.04.2014.

Abstract

Elliptic modular forms of weight 2 and elliptic modular curves are strongly related. In the rank-2 Drinfeld module situation, we have still modular curves that can be described analytically through Drinfeld modular forms. In [GR96] Gekeler and Reversat prove how the results of [Dri74] can be used to construct the analytic uniformization of the elliptic curve attached to a given automorphic form. In [Lon02] Longhi, building on ideas of Darmon, defines a multiplicative integral that theoretically allows to find the corresponding Tate parameter. In this thesis we develop and present a polynomial time algorithm to compute the integral proposed by Longhi. Also we devised a method to find a rational equation of the corresponding representative for the isogeny class.

Zusammenfassung

Elliptische Modulformen von Gewicht 2 und elliptische Modulkurven stehen in enger Verbindung. Im Fall eines Drinfeld-Moduls von Rang 2 haben wir noch Modulkurven, die durch Drifeldsche Modulformen analytisch beschrieben werden können. Gekeler und Reversat [GR96] beweisen, wie die Ergebnisse von [Dri74] genutzt werden können, um die analytische Uniformisierung der, einer gegeben automorphen Form eingeordneten elliptischen Kurve, zu konstruieren. Auf Darmons Ideen aufbauen definiert Lonhi [Lon02] ein multiplikatives Inte- gral, das es erlaubt, den entsprechenden Tate-Parameter zu finden. In der vor- liegenden Arbeit wird ein Polynomialzeitalgorithmus entwickelt und vorgestellt, um das von Longhi vogeschlagene Integral zu berechnen. Ausserdem wird eine Methode entwickelt, mit der eine rationale Gleichumg der entsprechenden Vertreter der Isogenieklasse gefunden werden kann.

Contents

Nomenclature vi

1 Introduction 1

2 Background 7

2.1 Notation...... 7

2.2 Notionsfromgraphtheory...... 7

2.3 TheDrinfeldupperhalfplane ...... 10

2.4 TheBruhat-Titstree ...... 11

2.5 Endsofthetree...... 13

2.6 Thereductionmap ...... 17

2.7 Drinfeldmodularcurves ...... 19

2.8 Thequotientgraph...... 20

2.9 Harmoniccocycles ...... 21

2.10 Thetafunctionsforarithmeticgroups ...... 24

2.10.1 Theta functions and harmonic cocycles ...... 25

iii Contents

2.11Heckeoperators...... 26

2.12 Application to the Shimura-Taniyama-Weil uniformization ...... 27

3 Integration, Theta function and uniformizations 31

3.1 Integration...... 31

3.1.1 Measuresandharmoniccocycles...... 32

3.1.2 The integral over ∂Ω...... 33

3.1.3 Changeofvariables...... 33

3.2 Thetafunction ...... 34

3.3 Complexuniformization ...... 35

3.4 p-adicUniformization...... 38

4 The Algorithm 45

4.1 Motivation...... 45

4.2 Elementaryfunctions...... 46

4.3 AHeckeoperator ...... 50

4.4 The change of variables and calculation of the integral ...... 61

4.4.1 Choosing the z0 ...... 62

4.4.2 Thepartitionoftheborder ...... 62

4.4.3 Thechangeofvariables...... 63

5 Applications and examples 69

5.1 Ellipticcurves...... 69

5.2 Supersingular elliptic curves ...... 78

5.3 Elliptic curves over C andEisensteinseries ...... 79

iv Contents

5.4 Reduction modulo p ofmodularforms ...... 84

5.5 TheTateCurve...... 87

5.6 ObtainingtheTateparameter ...... 90

5.7 Obtainingequationsforthecurves ...... 91

5.7.1 Elliptic curves in characteristic 2 and 3 ...... 92

5.7.2 Elliptic curves over characteristic p> 3...... 102

A Algorithms for the Quotient graph 113

A.1 Computational complexity of mathematical operations ...... 113

A.2 Representatives for the edges of Γ T ...... 115 0 \ A.3 Lifting cycles to T ...... 124

A.4 Findingtherepresentative ...... 126

B Algorithms for the table 129

B.1 Algorithms for the calculation of the table ...... 129

B.2 Algorithms for the calculation of the integral ...... 141

C Tables 149

C.1 Preliminaries ...... 149

C.2 Table for degree 3 over F3 ...... 150

C.3 Table for degree 4 over F3 ...... 150

C.4 Table for degree 3 over F5 ...... 152

C.5 Table for non-primes of degree 4 over F5 ...... 153

C.6 Table for primes of degree 4 over F5 ...... 166

C.7 Table for primes of degree 5 over F5 ...... 168

v Contents

References 174

vi 1. Introduction

The theory of modular forms and their relation to the arithmetic of elliptic curves is a central subject in modern mathematics, where most diverse branches of mathematics come together: complex analysis, algebraic geometry, representation theory, algebra and number theory. One of the most exciting and widely mathematical discoveries is the proof, by Andrew Wiles, of “Fermat last’s theorem”, its solution draws an incredible range of modern mathematics, which is precisely the relation between modular forms and elliptic curves.

There are a number of analogies between on the one hand, the integers Z and the rational numbers Q and on the other hand, Fq[T ] and its field of fractions Fq(T ). Frequently a problem posed in number fields or, in other words, in finite extensions of Q, admits an analogous problem in function fields, and the other way around. For example, since the appearance of Drinfeld’s work [Dri74], we know that all elliptic curves which are semistable at the place are modular, that is, they appear as a factor of the Jacobian of a Drinfeld ∞ modular curve. We are interested in number theory over function fields, particularly in

elliptic curves over Fq(T ).

In order to get a better idea of the function ﬁeld case, it is worth to start with a short description of the classical case, that is, over the rational numbers Q. Let f : H C be −→ a cuspidal modular form of weight two for the Hecke congruence subgroup Γ0(N) which is also a new eigenform with rational Hecke eigenvalues, we call it for short “a Q-rational newform” of level N. From the Eichler-Shimura theory, with f one can associate an elliptic curve E with conductor N and a morphism deﬁned over Q form the modular curve X0(N) to the elliptic curve E. Such an elliptic curve E is called a Weil curve.

On the other hand, since the 60’s (after the work of Shimura, Taniyama, and Weil) emerged

1 1. Introduction the conjecture that all elliptic curves over Q (up to isogeny) should be obtainable from the Eichler-Shimura construction. The conjecture known as the Shimura-Taniyama-Weil conjecture, is now a theorem called the modularity theorem [BCDT]. Basically it states that there are canonical bijections between the sets of

1) Normalized Q-rational newforms f of level N with rational Hecke eigenvalues;

2) One dimensional isogeny factors of the new part of the Jacobian of the modular curve

X0(N);

3) Isogeny classes of elliptic curves E over Q with conductor N.

The previous correspondence yields an eﬀective method to determine all elliptic curves E/Q with a given conductor N, since the modular parametrization is explicitly and eﬀectively computable (c.f. [Cre97]). For tables and numerical results see ([Ibid., Ch. 4]).

For some applications it is convenient to consider other kind of parametrizations instead of the modular one, for example the Shimura parametrization, introdued in [BC91] and [BD98]. Let E be an elliptic curve of conductor N and suppose that N is square free and + factorizes as N = N −N where N − has a even number of factors. Then there exits a parametrization of E by the Shimura curve X − + (cf. 3.4), that is a non constant mor- N N § phism from the Jacobian of the Shimura curve to E. However the lack of q-expansions for modular forms on non-split quaternion algebras, forces one to consider p-adic uniformizations of E by certain discrete arithmetic subgroups of SL2(Qp) at the primes p dividing

N −.

The modular forms considered here may be regarded as functions on oriented edges of certain Bruhat-Tits tree T (cf. 3.4), called harmonic cocycles, which can be identified § 1 with measures on P (Qp). Using these measures Bertolini and Darmon are able to define 1 a multiplicative p-adic integral defined over P (Qp), which theoretically gives rise to the modular parametrization of the Tate curve attached to a given harmonic cocycle.

In [Gre06], Greenberg gives an algorithm, running in polynomial time, for evaluating this p-adic integral up to a given precision. The key of his algorithm is a method devised by Pollack and Stevens [PS11] for explicitly lifting standard modular symbols to overconvergent ones. Although Greenberg is able to compute with good accuracy the Tate parameter, he does not give an explicit method to ﬁnd an equation deﬁded over Q for the elliptic curve

2 1. Introduction

as Cremona does. However in a recent preprint [GMS] Guitart et al, get p-adic approx- imations to the algebraic invariants of the elliptic curve, allowing for the recovery of the Weierstrass equation.

In the function field case, we want to describe a similar relationship between elliptic curves and automorphic forms. Although this result was proved for more general global fields (cf. [GR96]), we consider here the case where Q is replaced by the rational function field

K = Fq(T ).

The automorphic forms considered, may be regarded also as functions on the oriented edges of the Bruhat-Tits tree for GL2(K ), where K is the completion of K at the place ∞ ∞ = 1/T . In analogy with the classical case, in [GR96] Gekeler and Reversat prove that ∞ every elliptic curve E over F (T ) with conductor n , where n is an ideal of F [T ], is q ∞ q isogenous to a factor of the Jacobian of the Drinfeld modular curve M0(n) or equivalently E is a Weil curve. One can also establish a canonical bijection between the sets of:

1) Q rational new eigencocycles of level n with rational eigenvalues;

2) Isogeny factors of dimension one of the Jacobian of the Drinfeld modular curve M0(n);

3) K-isogeny classes of elliptic curves E/K with conductor n . ∞

This modular parametrization is explicitly constructed using a Theta function, however, this construction requires to calculate certain infinite product (cf. (2.6)), which makes it computationally hard to find the Tate parameter. On the other hand, Longhi [Lon02] working on function field analogues of Bertolini-Darmon [BD98], defines a multiplicative integral over P1(K ) and constructs a theta function in a different way as the one of ∞ Gekeler and Reversat. This approach does not give either an explicit method to calculate such integral.

In this thesis we develop an effective method to compute the multiplier of Gekeler’s theta function using the integral proposed by Longhi. We are able to calculate the Tate parameter q up to an accuracy of πM in running time O(M 7) operations; this is the main result of this thesis (c.f. Thm. B.2.1). In contrast to Greenberg‘s work, we go even further and find the corresponding elliptic curve defined over Fq(T ) in the isogeny class of the Tate curve corresponding to q. The importance of the these results, is that we now have a method to

3 1. Introduction

construct, at least for small primes and polynomials N of small degree, tables as Cremona does in the classical case.

We start with a harmonic cocycle ϕ as above, that is, a new eigencocycle of level n with 1 rational Hecke eigenvalues. From 3.1.1 we know that there is a measure µϕ on P (K ) § ∞ associated to ϕ. We use it, following Longhi, to deﬁne a multiplicative integral (3.9) which is very similar to the one used by Greenberg in the p-adic uniformization.

Motivated by the work of Greenberg, Darmon and pollack, we develop an algorithm to calculate our integral up to a given fixed precision of M digits. A crucial tool in Greenberg’s calculation is the logarithm, which allows him to pass from a multiplicative integral to an additive one. In the function field case we had to replace the use of logarithm by a different

method which lead us to calculate several integrals of functions in the ring 1 + πFqJπ, tK deﬁned on O (cf. (4.3)), instead of P(K ) (cf. (4.1)). This idea and the concepts involved ∞ ∞ were sketched to me by Prof. B¨ockle.

In order to calculate the integral we deﬁne a Hecke operator over certain set S (cf. (4.7)) of functions which are I -equivariant. We show that the integral over any edge of the ∞ tree T can be regarded as an element of S and they are eigen-functions with eigenvalue 1. The last property allows us to calculate the integral over all edges in T and functions on M 1+ πFqJπ, tK modulo π .

The calculation of the Hecke operator U requires to work with the edges of the quotient ∞ graph Γ0(N) GL2(K )/I , where N is a polynomial in Fq[T ] that generates the ideal \ ∞ ∞ n, which is a covering of double the quotient GL2(A) GL2(K )/I . Since most of the \ ∞ ∞ algorithms available in the literature are made to work with the vertices of the quotient graph, we need to describe sets of representatives for the edges of the quotient graphs above and implement algorithms that allow us to work with such as sets (c.f Appendix A). The algorithms and the proofs that appear in the Appendices A and B, were made with the help of Dr. Cervi˜no.

Once we have the Tate parameter q we proceed to ﬁnd a representative of the elliptic

curve in the isogeny class deﬁned over Fq(T ). Unfortunately the coeﬃcients a4(q) and

a6(q) of the Tate curve ( 5.12) are not rational (cf. Example 5.7.4). So we need to ﬁnd an appropriate change of variables to transform the Tate curve into a rational model.

In the case of characteristic 2 and 3 a simple change of variables to transform the Tate

4 1. Introduction

curve was enough to find the rational model. In characteristic greater than 3 we need to consider the Eisenstein form y2 = 4x3 g x g of the curve. Following a suggestion of − 2 − 3 Prof. Böckle, we prove that there are powers of g2 and g3 that are rational functions (cf. Proposition 5.7.9). Hence after an appropriate change of variables we get a model defined over the field Fp(T ). A direct consequence of this is that the Eisenstein series are algebraic over Fq(T ).

This thesis is organized as follows:

In chapter 2 we give all the preliminaries from the article of Gekeler-Reversat [GR96], which include a short introduction to graphs in 2.2. We give the basic definitions of the § Drinfeld upper half plane, Bruhat-Tits trees, ends, the reduction map, the quotient graph and harmonic cocycles in sections 2.3-2.9. In Section 2.5, where the ends of the tree T §§ are defined, we explain how they are identified with P1(K ) (cf. Lemma 2.5.4) and how ∞ each edge induces a partition of P1(K ) into two disjoint open sets. Besides the definion ∞ of the harmonic cocycles in 2.9 we show how they can be constructed using the homology § of the quotient graph (cf. Lemma 2.9.4).

The construction of Gekeler and Reversat of theta series is given in 2.10. Theorem 2.10.2 § gives their main properties. There it is stated that the multiplier of the theta series induces a symmetric bilinear pairing, this is used latter to find the integral with the minimal valuation, which is precisely the Tate parameter. Theorem 2.10.4, mostly due to Van der Put, gives the relation between theta functions and the harmonic cocycles. This relation allows us latter to construct the theta series by means of a multiplicative integral. We finish this chapter with the definition of the Hecke operator in 2.11 and the applications § to the Shimura-Taniyama-Weil uniformisation 2.12, which is the main result of [GR96]. § In Chapter 3 the definition of Longhi’s multiplicative integral (cf. Def. 3.1) over any compact X is given. In 3.1.1 we state the relation between harmonic cocycles and mea- § sures, which is a direct consequence of Lemma 2.5.4. In 3.1.2 we give the definition of § the multiplicative integral when X is P1(K ) and the measure is the one induced from the ∞ harmonic cocycle. The theta function as a multiplicative integral is the content of Theorem 3.2.1, which allows us to show that the multiplier of the theta function can be given as a multiplicative integral (cf. (3.9)). This is the integral that we are interested in, since it allows us to determine the Tate parameter. This chapter also includes two sections with

5 1. Introduction

the classical complex uniformisation in 3.3 and the p-adic uniformisation in 3.4 with a § § short explanation of Greenberg’s algorithm.

In Chapter 4 we explain our algorithm to calculate the integral. This is the main theoretical contribution of this thesis. In 4.2 we define F , the set of fundamental functions. We show § I in Lemma 4.2.1 that FI is a group and we define the action of the Iwahori subgroup on it (cf. (4.6)). We introduce in 4.3 the Hecke operator U and the set S (4.7) on which § ∞ the operator U acts. The main results of this section are Lemma 4.3.7 and Proposition ∞ 4.3.8 in which we prove that the integrals regarded as elements of S are I -equivariant ∞ and eigenfunctions of the operator U , respectively. At end of section 4.3 we explain how ∞ § to calculate the Table by applying the Hecke operator U . This section finishes with the ∞ example 4.3.13 of how to apply the operator U . ∞ As we already explained we need to transform our integral (3.9) into one of the form (4.1) defined over O . In 4.4 we explain how to carry out this change of variables and we ∞ § perform carefully all the computations over all possible open sets arising from the partition of the border P1(K ). ∞ Chapter 5 deals with the applications of our algorithm to calculate the integral (3.9). We start this chapter with some definitions and results on elliptic curves, modular forms and the Tate curve (cf. 5.1-5.5). We explain in 5.6 how to calculate the Tate parameter. §§ § This includes a calculation of the valuation of q and Theorem 5.6.1 which states that we can compute the Tate parameter q up to accuracy πM in time O(M 9).

In 5.7.1 we explain how to find the rational model for the Tate curve in characteristic 2 § and 3 and write the corresponding algorithm and give some examples. Section 5.7.2 deals § with characteristic p> 3. We show first the rationality or certain powers of the Eisenstein series g2 and g3 (cf. Proposition 5.7.9) over the field Fq(q). We also give the algorithm to find curves in characteristic p> 3 and we give examples for p =5, 7, 11 and 13.

In the appendices A and B we explain the implementations for algorithms to deal with the quotient graph and the table respectively, as well as the running time for the main algorithms and the proof of Theorem 5.6.1. Appendix C includes tables for some primes and small degree of n.

Implementations of the algorithms described in this thesis, were done on the algebra system Magma [BCP97] and are available upon request.

6 2. Background

2.1 Notation

We recall some facts on the Drinfeld upper half plane, the Bruhat-Tits tree, harmonic cocycles and theta functions1. We ﬁx throughout this work the following notation, assuming the reader to be familiar with [GR96].

Fq = the finite field of characteristic p with q elements A = Fq[T ] K = Fq(T ) = the fixed place of K of degree one corresponding to v ∞ 1/T val = the valuation v1/T 1 K = Fq((π)), π = T − ∞ O = Fq[[π]], the -adic integers ∞ ∞ C = the completion of an algebraic closure of K ∞ ∞ val( ) = the multiplicative norm on C that extents q · : K R>0. |·| ∞ →

2.2 Notions from graph theory

Deﬁnition 2.2.1. Let S be a non empty countable set.

(a) A (directed multi-)graph G is a pair (X(G), Y(G)) where X(G) is a (possibly inﬁnite) non-empty set and Y(G) is a subset of X(G) X(G) S such that × ×

1. if e =(v, v′,s) lies in Y(G), then so does its oppositee ¯ =(v′,v,s),

1The theta functions considered here are the rigid analytic functions deﬁned in [GR96, 5] §

7 2. Background

2. for any (v, v′) X(G) X(G), the set s S (v, v′,s) Y(G) is a ﬁnite set ∈ × { ∈ | ∈ } whose cardinality is denoted by nv,v′ ,

3. for any v X(G), the set Nbs(v) := v′ X(G) (v, v′,s) Y(G) for some s ∈ { ∈ | ∈ ∈ S is ﬁnite. }

(b) A subgraph G′ G is a graph G′ such that X(G′) X(G) and Y(G′) Y(G). ⊂ ⊂ ⊂

Notation 2.2.2. An element v X(G) is called a vertex and an element e Y(G) is called ∈ ∈ an oriented edge. If the cardinality of S is one, we simply write (v, v′) instead of (v, v′,s).

Deﬁnition 2.2.3. (a) For each edge e =(v, v′,s) Y(G) we call o(e) := v the origin of ∈ e and t(e) := v′ the target of e.

(b) Two vertices v, v′ are called adjacent, if v, v′ = o(e), t(e) for some edge e. { } { } Deﬁnition 2.2.4. An edge e with o(e)= t(e) is called a loop. A vertex v with # Nbs(v)=1 is called terminal.

Deﬁnition 2.2.5. Assume G to be a graph.

(a) Let v, v′ X(G). A path ω from v to v′ is a ﬁnite subset e ,...,e of Y(G) such ∈ { 1 k} that t(e )= o(e ) for all i =1,...,k 1 and o(e )= v, t(e )= v′. i i+1 − 1 k (b) The length of a path ω is the number of edges contained in it.

from v to v′ (or if no such path exists). ∞

(d) A path e ,...,e from v to v′ without backtracking, i.e., such that for no i we have { 1 k} ei =e ¯i 1, is called a geodesic. −

Note that the length of a geodesic need not be d(v, v′) but that d(v, v′) is attained for a geodesic.

Example 2.2.6. Consider the set S = and the set of vertices {∗} X(G)= 1, 2, ..., 16 . { }

8 2.2. Notions from graph theory

If the graph G is represented by

6 10 14 ?>=<89:; GFED@ABC GFED@ABC

?>=<89:;1 ❃ ❃❃ ❃❃ ❃❃ ❃❃ 4 ?>=<89:;✴ ✴✴ ✴ ✴✴ ✴ ✴✴ ?>=<89:;2 ❃ ?>=<89:;7 ✴ GFED@ABC11 GFED@ABC15 ❃❃ ✴✴ ❃❃ ✴ ❃❃ ✴✴ ❃❃ ✴ 5 9 13 ?>=<89:; ?>=<89:; GFED@ABC

?>=<89:;3 ❃ ❃❃ ❃❃ ❃❃ ❃❃ ?>=<89:;8 GFED@ABC12 GFED@ABC16 ,

then the set of edges is

(1, 6), (1, 4), (2, 4), (2, 7), (2, 5), (3, 5), (3, 8), (6, 10), (4, 9), { (7, 11), (5, 9), (8, 12), (10, 14), (11, 15), (9, 13), (12, 16) ; } here due to Deﬁnition 2.2.1 only one of (v, v′) (or (v′, v)) is written since either both are elements of Y (G) or none.

A graph G is connected if for any two vertices v, v′ X(G) there is a path from v to v′. A ∈ cycle of G is a geodesic from some vertex v to itself. Therefore a loop is a cycle of length one. A graph G is cycle-free if it contains no cycles.

Deﬁnition 2.2.7. A graph G is called a tree if it is connected and cycle-free.

If G is a tree, then any two vertices of G are connected by a unique geodesic. Any subgraph T G which is a tree is called subtree. A maximal subtree in a graph G is a subtree which ⊆ is maximal under inclusion among all subtrees of G.

9 2. Background

Definition 2.2.8. The degree of v X(G) is ∈ deg(v) := # e Y (G) o(e)= v . { ∈ | } Thus v is terminal precisely if deg(v) = 1. A graph G is called k-regular if for all vertices v X(G) we have deg(v)= k. ∈ A graph G is finite if # X(G) < . Then also # Y(G) < , since deg(v) is finite for all ∞ ∞ v X(G). The diameter of a (finite) graph G is ∈ diam(G) := max d(v, v′). v,v′ X(G) ∈

2.3 The Drinfeld upper half plane

The Drinfeld upper half plane is the set Ω := P1(C ) P1(K ) ∞ \ ∞ on which GL2(K ) acts through fractional linear transformations by ∞ az + b γ z := , h i cz + d a b where γ = ( c d ) GL2(K ) and z Ω. This action is well defined since GL2(K ) ∈ ∞ ∈ ∞ preserves K . We define the boundary of Ω, denoted by ∂Ω, to be P1(K ). ∞ ∞ In order to define the algebra of rigid analytic functions on Ω, we consider the following sets that form an admissible cover of Ω (cf. [FvdP81, Ch. II] for details). Define the imaginary absolute value on Ω as the distance from K : ∞

z i := inf z x x K . | | {| − || ∈ ∞} Deﬁnition 2.3.1. Let z Ω and n N, a basic aﬃnoid is a set of the form ∈ ∈ n n A := z Ω q− 6 z , z 6 q . n ∈ | | |i | | A function f : An C is holomorphic if it is the uniform limit of rational functions → ∞ without poles in An. The collection of such functions is denoted by OΩ(An). With the norm

f A := sup f(z) z A , k k n {| || ∈ n}

the space OΩ(An) is a Banach algebra and OΩ(Ω) := lim OΩ(An) is a Fr´echet space [FvdP81, Ch. III]. ←−

10 2.4. The Bruhat-Tits tree

2.4 The Bruhat-Tits tree

In this section we recall the deﬁnition of a graph T called the Bruhat-Tits tree of P GL2(K ), ∞ which is a basic combinatorial object for the arithmetic of K . For more details see [Ser03, ∞ Ch. II].

2 A lattice in K is a free O -submodule of rank 2. We say that two lattices L1 and L2 are ∞ ∞ homothetic if L1 is a K× -multiple of L2. Homothety is an equivalence relation. The set ∞ of vertices X(T) of T consists of the homothety classes of O -lattices. Two vertices are ∞ joined by an edge if and only if they can be represented by lattices L1, L2 such that there

are strict inclusions πL2 L1 L2. The set of oriented edges of T is denoted by Y (T).

Thus the vertices and edges of T may be described as follows:

Classes [L1] of O -lattices L1 X(T) = Vertices of T = ∞ , { } in K2 ∞

Ordered pairs ([L1], [L2]) with representatives Y (T) = Oriented edges of T = L , L such that L L and . { }  1 2 1 ⊂ 2   with πL2 L1 L2    It is well known that the graph T is a connected regular tree of degree q + 1, where q is the cardinality of the residue ﬁeld of K [Ser03, Ch. II. Thm. 1 ]. ∞

The group GL2(K ) acts naturally on lattice classes by left multiplication (g, [L]) [gL]. ∞ 7→ This induces a transitive action on the vertices of T which preserves the incidence relations

πL2 L1 L2. In this way one obtain an action of GL2(K ) on T. ∞ i For i Z let vi be the vertex [π O O ] of T. Since the vertex v0 = [O O ] has ∈ ∞ ⊕ ∞ ∞ ⊕ ∞ stabilizer K× GL2(O ) in GL2(K ), we have the following ∞ ∞ ∞ Proposition 2.4.1. There is a canonical bijection

GL2(K )/K× GL2(O ) X(T) ∞ ∞ ∞ −→ γ K× GL2(O ) γ[O O ], · ∞ ∞ 7−→ ∞ ⊕ ∞ equivariant for the left action of GL2(K ). ∞

11 2. Background

Deﬁnition 2.4.2. We deﬁne the Iwahori subgroup of GL2(K ) as ∞

a b I := GL2(O ) such that c 0(mod π) . (2.1) ( c d ! ∈ ∞ ≡ )

Similarly, we label the edges e (i Z) such that t(e )= o(e )= v . Each edge is taken i ∈ i i+1 i by GL2(K ) to e0 = (v 1, v0), so it follows that Y (T) = GL2(K )/StabGL2(K∞)(e0). A ∞ − ∼ ∞ simple computation shows StabGL2(K∞)(e0)= K× I. ∞ Proposition 2.4.3. There is a canonical bijection

GL2(K )/K× I Y (T) ∞ ∞ −→ 1 γ K× I (γ[π− O O ],γ[O O ]), · ∞ 7−→ ∞ ⊕ ∞ ∞ ⊕ ∞

equivariant for the left action of GL2(K ). ∞

From now on given a γ GL2(K ) we denote its class in GL2(K )/K× GL2(O ) as ∈ ∞ ∞ ∞ ∞ [γ]0 and its class in GL2(K )/K× I as [γ]1. We may implicitly use by abuse of nota- ∞ ∞ tion the bijections above and understand [γ]0 and [γ]1 as vertex and edge respectively.

The vertex v0 = [O O ] is called the standard vertex, analogously the edge e0 = ∞ ⊕ ∞ 1 ([π− O O ], [O O ]) is called the standard edge. ∞ ⊕ ∞ ∞ ⊕ ∞ The next two lemmas will help us to identify the vertices of the tree with explicitly given matrices. Furthermore, they will show which matrices correspond to adjacent vertices in the tree.

Lemma 2.4.4. Every class of GL2(K )/K× GL2(O ) has a unique representative of the ∞ ∞ ∞ form πn u 0 1! with n Z and u K /πnO . ∈ ∈ ∞ ∞

One can ﬁnd a constructive proof in [But12, Lemma 2.7]. We call this representative the vertex normal form. In what follows we will denote the vertex represented by the matrix πk u in normal form 0 1 as [k,u].

12 2.5. Ends of the tree

Lemma 2.4.5. Consider the two matrices in vertex normal form

πn u πn+1 u + aπn A := , B := 0 1! 0 1 !

n with n Z, a Fq, u K /π O and let L1 and L2 be the two lattices ∈ ∈ ∈ ∞ ∞ 2 2 L1 := AO , L2 := BO . ∞ ∞ Then L L and L /L = F . 1 ⊃ 2 1 2 ∼ q

Recall that T is a regular tree of degree q +1. The previous lemma only displays q vertices − − πn 1 u mod πn 1O∞ 2 adjacent to [L1]. The remaining one is the class of 0 1 O . ∞ The subgroup I is not normal in GL2(K ). Denoting by N the normalizer of I in ∞ 0 1 GL2(K ), we have that N/IK = Z/2. As one can easily see, δ := ( π 0 ) is a repre- ∞ ∞ ∼ sentative of the non-trivial quotient class of N/I. Let γ GL2(K ) such that [γ]1 = e, ∈ ∞ then multiplication from the right with δ corresponds to the map

Y (T) Y (T) −→ e e¯ 7−→ that is [γδ]1 =e ¯.

There are two canonical projection maps from X(T) X(T) S to X(T), which induce × × two maps from Y (T) to X(T). We choose the ﬁrst projection map that associates to each e its origin o(e), i.e.,

pr : Y (T) X(T) (2.2) 1 −→ e =(v, v′) o(e)= v . (2.3) 7−→

This map is also compatible with multiplication by δ that is, pr1(δe)= o(δe)= t(e).

2.5 Ends of the tree

Before relating the Bruhat-Tits tree to the Drinfeld upper half plane, we need to deﬁne the ends of the tree.

2 Let e1, e2 the standard basis of K as a column vector. { } ∞

13 2. Background

Deﬁnition 2.5.1. Let ([L0], [L1], ...) and ([L0′ ], [L1′ ], ...) be two non-backtracking inﬁnite

sequences of adjacent vertices. We say that they are equivalent if [Ln] = [Ln′ +m] for some m Z and all n large enough. An end of T is an equivalence class of such sequences. The ∈ collection of ends of T is denoted by Ends(T).

From the elementary divisor theorem we obtain the following. (cf. [Bos09, Ch. 6] for details).

2 Proposition 2.5.2. Let L and L′ be two lattices in K . Then there exists a O -basis ∞ ∞ a b e1, e2 of L and integers a, b such that π e1, π e2 is a O -basis of L′. The numbers a { } { } ∞ and b are independent of the choice of the basis of L and L′.

Remark 2.5.3. i) If L′ L then the numbers a and b from the previous proposition ⊂ are positive. Furthermore, we can ﬁnd in the class [L′] a lattice L′′, namely L′′ = min a,b L′π− { } such that if f , f is a basis of L then L′′ is generated by one of the f { 1 2} i and the other one multiplied by a positive power of π.

ii) Given an end s there is a unique representative sequence starting with the lattice 2 L0 = O e1 O e2, were ei is the standard basis of K . ∞ ⊕ ∞ ∞

Lemma 2.5.4. There is a canonical GL2(K )-equivariant bijection ∞

φ : Ends(T) P1(K ). → ∞

Proof. Given s Ends(T) represented by the sequence ([L ], [L ], ...) we can construct a ∈ 0 1 representing sequence of lattices

L L L ... 0 ⊃ 1 ⊃ 2 ⊃ such that Ln/Ln+1 = O /πO for all n and πL0 does not contain any of the Ln’s, since ∼ ∞ ∞ 2 there is no backtracking. Therefore nLn is a O -submodule of K spanning a K -line. ∩ ∞ ∞ ∞ For any n, by the Remark 2.5.3, there exists a basis x ,y of L such that x , πny { n n} 0 { n n} is a basis of Ln, analogously for Ln+1. Since Ln+1 Ln we can write xn+1 in terms of n ⊂ the basis of Ln, that is xn+1 = xnan + π bnyn for some an, bn O and an πO . ∞ ∞ n ∈ 6∈ Then xn+1 xnan = π bnyn and since an πO we may replace xn+1 by anxn+1 to get − 6∈ ∞ x x (mod πn). Continuing in this way, we can construct a Cauchy sequence (x ) n+1 ≡ n n n>1 x1 2 which converges to a nonzero element x =( x2 ) of nLn K . ∩ ⊂ ∞

14 2.5. Ends of the tree

The bijection

φ : Ends(T) P(K2 ) = P1(K ) → ∞ ∼ ∞ e x + e x (x : x ) h 1 1 2 2i 7→ 1 2 2 is established by associating to the end s = ([Ln]) the line in K generated by nLn, that ∞ 2 ∩ is φ(s)= O (e1x1 + e2x2), where e1, e2 is the standard basis of K as column vectors. ∞ { } ∞ 2 x1 The map φ is surjective. Let l be a line in K generated by x = ( x2 ), that is l = ∞ n O (x1e1 + x2e2). Deﬁne the sequence of lattices Ln = O x π yO with y a vector in ∞ ∞ ∞ 2 ⊕ K linearly independent of x. Let s be ([Ln])n>1, clearly Ln+1 Ln and φ(s)= l. ∞ ⊂

The map φ is injective. Let s and s′ be two ends and let L and L′ be two sequences { n} { n} of lattices representing s and s′, respectively, whose intersection generate the same line. 2 Then up to multiplication by an scalar Ln = Ln′ = O x for some x K . Eliminating, ∩ ∩ ∞ ∈ ∞ if necessary, the ﬁrst terms of the sequences, one can assume that L0 = O x O y and ∞ ∞ n ⊕ L0′ = O x O y′ . Since x Ln for all n and [L0 : Ln]= q we have that Ln is generated ∞ ∞ n ⊕ ∈ n k by x, π y and similarly Ln′ is generated by x, π y′ . Since π y′ = ax + by for a, b O ∞ { } { } m ∈ for k large enough one ﬁnds that Lk′ +m is generated by x and π bw and therefore coincides with Lj+m for some j, that is the ends ([Ln]) and ([Ln′ ]) are equal, hence φ is injective.

x1 a b Finally, the map φ is equivariant. If L is generated by the vector ( x ), for a γ =( ) ∩n n 2 c d ∈ x1 ax1+bx2 GL2(K ) the generator of γLn is γ ( x2 )= . ∞ ∩ cx1+dx2 1 On the other hand GL2(K ) acts on P (K ) by M¨obius transformations, that is ∞ ∞ γ(x1 : x2)=(ax1 + bx2 : cx1 + dx2) and the equivariance follows.

Remark 2.5.5. In [GR96] the elements of K2 are taken as row vectors and the action ∞ 1 of GL2(K ) on the lattices is given by right multiplication Lγ− . So if Ln is gener- ∞ ∩ a b ated by the row vector (x1, x2) and γ = ( c d ) GL2(K ) then γLn is generated by ∈ ∞ ∩ (x d x c, x b + x a). Which does not correspond to γ(x : x )=(ax +bx : cx +dx ). 1 − 2 − 1 2 1 2 1 2 1 2 In order to make the map φ compatible with the action of GL2(K ), it needs to be com- ∞ posed with the canonical map (x1 : x2) ( x2 : x1), that is the M¨obius transformation 1 1 7→ − z z− on P (K ) (cf. [GR96, 1.3.2]). 7→ − ∞ §

From the bijection in Lemma 2.5.4, we have in particular that

the image of the semi-line whose vertices are represented by [k, 0] is • { }k<0

15 2. Background

= (1 : 0) P1(K ). To see this note that the line represented by [k, 0] is ∞ ∈ ∞ { }k<0 in the class of the sequence of lattices Lk whose basis are the columns of the matrix 0 1 πk 0 . the semi-line whose vertices are represented by [k, 0] goes to • { }k>0 0=(0:1) P1(K ). ∈ ∞

We will denote by A(0, ) the “line” whose vertices are represented by [k, 0] k Z. ∞ { } ∈

From the normalization the edge e0 may be identiﬁed with the compact open set Ue0 = 1 O P (K ) by considering the ends passing through e0. These are represented by ∞ ⊂ ∞ πn u sequences of lattices whose basis are the columns of the matrix ( 0 1 ) with n > 1 and u K /πnO with positive valuation. This extends by equivariance to an assignment of ∈ ∞ ∞ a compact open subset of P1(K ) to each oriented edge of the tree by ∞

1 Uγe0 := γ O P (K ) γ GL2(K ). (2.4) h ∞i ⊂ ∞ ∀ ∈ ∞

1 1 Since the action of GL2(K ) on the disks of P (K ) is transitive, every disk of P (K ) is ∞ ∞ ∞ the image of an edge on Y (T). We observe some essential properties of the assignment:

i) Given any oriented edge e, then for the opposite edgee ¯, the associated open set 1 satisﬁes Ue¯ = P (K ) Ue. ∞ − ii) For any vertex v the set U , where e runs over all edges e with initial vertex v, { e} form a disjoint covering of P1(K ). ∞

1 iii) The sets Ue form a basis of compact open subsets of P (K ). { } ∞

Given two vertices v and v′ on the tree, there is a unique oriented path ω from v to v′ joining them. The open set associated to these two vertices is the union of all ends containing ω. Given an edge e and a vertex v, we say that e points away from v if the unique path with origin v and containing o(e) and t(e) contains e (and note ¯).

Remark 2.5.6. The end deﬁnes an “orientation” of the tree T, that is a decomposition + ∞ + of Y (T) = Y (T) ˙ Y −(T), where an edge e =(v , v ) belongs to Y (T) if it points to ∪ 1 2 ∞ and it is called positive and negative (e Y −(T)) otherwise. From this we get a section ∈

16 2.6. The reduction map

(X(T) ∼= Y +(T) ֒ Y (T −→ → v e 7−→ such that o(e)= v and e is positive.

+ Note that multiplication by δ allows us to change from Y (T) to Y −(T) and the other way around. In terms of matrices, we have that the map e e¯ is given by [g] [gδ] for a 7→ 1 7→ 1 πk u g GL2(K ). Then each edge e of Y (T) is uniquely represented by either (if e is ∈ ∞ 0 1 positive) or πk u ( 0 1 ) (if e is negative). 0 1 π 0

2.6 The reduction map

As constructed so far, the Bruhat-Tits tree T is a combinatorial object. Next we will associate to it a geometrical object. For this, we identify each edge with a copy of the real unit interval endowed with the usual topology. Then we glue edges according to the relations on T using the quotient topology and we denote by T(R) this new tree and it is called the geometrical realization of T . Let e be an edge of T(R) joining the vertices [L0] and [L ], any point on e is determined by its barycentric coordinates, i.e., for t [0, 1] we 1 ∈ write x = (1 t)[L ]+ t[L ] to indicate that x is the point “at distance t from the vertex − 0 1 [L0] in the direction of [L1]”. Denote by T(Q) the Q-points of the geometrical realization of T deﬁned as t [0, 1] Q and also T(Z) to be the vertices of T. ∈ ∩ The geometric realization of the tree T parametrizes norms on the two dimensional vector space K2 , as is stated by Goldman and Iwahori [GI63], which allows us to deﬁne the ∞ reduction map.

Deﬁnition 2.6.1. A real non-archimedean norm on K2 is a map ν : K2 R which ∞ ∞ → satisﬁes

1. ν(x) > 0 and ν(x)=0 x =0, ⇔ 2. ν(ax)= a ν(x) for a K , | | ∈ ∞ 3. ν(x + y) 6 max ν(x), ν(y) . { }

17 2. Background

Given a lattice L on K2 , one can associate the norm ∞

νL(x) := inf a a K× , x aL . | | | ∈ ∞ ∈ We say that two norms ν and ν′ are similar if there is a constant t R such that ν = tν′. ∈ Moreover, the group GL2(K ) acts on the space of norms by (γν)(x) := ν(γx) for all ∞ γ GL2(K ) (cf. [GR96, 1.4.2]). ∈ ∞ § Theorem 2.6.2 (Goldman-Iwahori). There is a canonical bijection compatible with the

right action of GL2(K ) ∞ Similarity classes of real non-archimedean norms ν on K2 T(R). ∞ ←→ In particular,

Classes of norms whose unit ball is an O -lattice T(Z). { ∞ } ←→

2 For z Ω we deﬁne the norm νz on K as follows: ∈ ∞ 2 νz : K R 0 ∞ −→ ≥ (u, v) ν ((u, v)) := uz + v 7−→ z | | and the reduction map is deﬁned by

λ : Ω T(R) −→ z [ν ]. 7−→ z Q Since C× = q , it is not diﬃcult to prove the following (cf. [Gek99]): | ∞| Proposition 2.6.3. One has a surjection λ : Ω ։ T(Q).

The previous proposition yields the following well known properties of the reduction map:

1 1 λ− (vertex)= P (C× ) (q + 1) disjoint balls, in particular, • ∞ − 1 λ− (v0)= z C× z 1, z c 1 c Fq . • { ∈ ∞ || | ≤ | − | ≥ ∀ ∈ }

We will refer to last set as the standard aﬃnoid.

Remark 2.6.4. From the GL2(K )-equivariance of the reduction map we can in principle ∞ 1 1 ﬁnd the ﬁber of any vertex and any edge on the tree from λ− (v0) and λ− (e0).

18 2.7. Drinfeld modular curves

2.7 Drinfeld modular curves

An arithmetic subgroup Γ of GL2(K ) is a subgroup commensurable with Γ(1) := GL2(A) ∞ (cf. [Gek97, 3.1]), i.e., such that Γ Γ(1) has ﬁnite index in both Γ and Γ(1). The main § ∩ examples are the principal congruence subgroups deﬁned as follows.

Deﬁnition 2.7.1. Let N be any monic polynomial in A = Fq[T ] and consider the set

1 0 Γ(N)= γ GL (A) γ (mod N) . ∈ 2 ≡ ( 0 1 ! )

A subgroup of GL2(A) that contains Γ(N) is called a congruence subgroup. Special cases are

Γ0(N)= γ GL2(A) γ ∗ ∗ (mod N) ∈ ≡ 0 ( ! ) ∗ and

1 Γ (N)= γ GL (A) γ ∗ (mod N) . 1 ∈ 2 ≡ ( 0 1 ! )

Most of the properties stated in this chapter for arithmetic subgroups Γ are proved for Γ = Γ(1) in [Ser03] or in [Non01], and they can be proved for general arithmetic subgroups as deﬁned above.

Let Γ be a congruence subgroup, it acts on Ω by fractional linear transformations with ﬁnite stabilizers. The quotient Γ Ω is a rigid analytic space over K . Moreover, it is \ ∞ smooth of dimension one. In fact, the analytic curve Γ Ω can be shown to arise from an \ algebraic curve (cf. [Gek86, Ch. V]).

Theorem 2.7.2 (Drinfeld). There exists a smooth irreducible affine algebraic curve YΓ defined over C such that Γ Ω and the underlying analytic space are canonically isomorphic ∞ \ as analytic spaces over C . ∞ The curve Y can be compactified by adding the finite set of cusps Γ P1(K). We denote Γ \ this compactification by XΓ, that is

X =Γ Ω ˙ Γ P1(K). Γ \ ∪ \

The curves XΓ will be referred to as Drinfeld modular curves. IfΓ=Γ0(N) we write

X0(N) instead of XΓ.

19 2. Background

2.8 The quotient graph

As already mentioned, the group GL2(K ) acts on the tree. Since GL2(A) is discrete in ∞ GL2(K ), the stabilizers in GL2(A) of edges and vertices are finite. As is proved in [Ser03], ∞ the quotient GL (A) T is a “half line”. In particular, it is isomorphic to the subtree of T 2 \ whose vertices are represented by [k, 0] . { }k>0 As proved in [Non01] for Γ a congruence subgroup, the quotient Γ T is a connected \ graph which is the union of a finite graph with a finite number of “half lines” attached to it. These are in one-to-one correspondence with the cusps of the corresponding Drinfeld modular curve and will be henceforth called cusps.

For a congruence subgroup Γ the quotient graph Γ T is a “ramiﬁed covering” \ π :Γ T GL (A) T. \ −→ 2 \ Given any edge e T, we will denote bye ˜ its class on the quotient graph, analogously for ∈ any vertex v its class in the quotient is denoted byv ˜. Also we say that a vertex in Γ T \ has level i if its projection under the map π is Λi.

3 Observe that the ﬁgure below shows the quotient graph for Γ0(N) with N = T over F2.

6 10 / 14 ❴❴❴ ?>=<89:; GFED@ABC GFED@ABC

?>=<89:;1 ❃ ❃❃ ❃❃ ❃❃ ❃❃ 4 ?>=<89:;✴ ✴✴ ✴ ✴✴ ✴ ✴✴ ❴❴❴ ?>=<89:;2 ❃ ?>=<89:;7 ✴ GFED@ABC11 GFED@ABC15 ❃❃ ✴✴ ❃❃ ✴ ❃❃ ✴✴ ❃❃ ✴ 5 9 13 ❴❴❴ ?>=<89:; ?>=<89:; GFED@ABC

?>=<89:;3 ❃ ❃❃ ❃❃ ❃❃ ❃❃ ❴❴❴ ?>=<89:;8 GFED@ABC12 GFED@ABC16

20 2.9. Harmonic cocycles

is a covering of Λ Λ Λ Λ 0 1 2 3 ··· n where Λn = [π O O ]. ∞ ⊕ ∞

Denote by (Γ T)◦ the subgraph of Γ T obtained by removing all the edges starting at \ \ level deg(N) and the vertices starting at level deg(N)+1, (we know (cf. [Non01]) that at level deg(N) there is no more identiﬁcations between edges).

2.9 Harmonic cocycles

In this section Γ will be a congruence subgroup.

Deﬁnition 2.9.1. Let G be an abelian group. A G-valued harmonic cocycle on the tree T is a map ϕ : X(T) G that satisﬁes −→

i) ϕ(¯e)= ϕ(e) for all e Y (T), − ∈ ii) ϕ(e) = 0 for all v X(T). t(e)=v ∈ P The group of harmonic cocycles on T with values in G is denoted by H(T,G). We denote Γ by H!(T,G) the subspace of H(T,G) that also satisﬁes the following conditions

iii) ϕ(γe)= ϕ(e) for all γ Γ, ∈ iv) ϕ has compact support modulo Γ, i.e. modulo Γ, the set of edges were ϕ takes non-zero values is ﬁnite.

We are interested in Z-valued harmonic cocycles. From the property iii) the elements of Γ H!(T, Z) may be regarded as invariant functions on the oriented edges of the quotient ˜ ˜ graph Γ T. Let us denote by Γ := Γ/(Γ Z(K )) and Γt(˜e) := StabΓ˜ (t(e)) for some lift e \ ∩ ∞ to T ofe ˜, similarly we deﬁne Γ˜e˜ and the quantity m(˜e) := [Γ˜t(˜e) : Γ˜e˜]. The condition ii) is equivalent to the following sum condition (with multiplicities that count how many edges of T are identiﬁed modulo Γ):

m(˜e)ϕ(˜e) = 0 for allv ˜ X(Γ T). (2.5) ∈ \ t(˜Xe)=˜v

21 2. Background

Before constructing explicitly the space of harmonic cocycles, we need to recall deﬁnitions of some further groups. First the maximal torsion-free abelian quotient of Γ, namely ab ab Γ¯ := Γ /tor(Γ ). Also Γf will denote the normal subgroup generated by elements of ﬁnite order. This two groups are related by the following lemma, which holds for any congruence subgroup [Non01].

ab Lemma 2.9.2. The groups Γ¯ and (Γ/Γf) are isomorphic.

In [Ser03, Ch. I, Thm. 13] is proved that the group Γ/Γf is canonically identiﬁed with the fundamental group of the graph Γ T, so Γ/Γ is free (cf. [Ser03, p. 43]). The group \ f Γ¯ is isomorphic to the homotopy group of the quotient graph, indeed we have the following:

Lemma 2.9.3. Let v be a vertex of T, there exists an isomorphism

i : Γ¯ H (Γ T, Z) → 1 \ α ψ 7−→ α,v

where ψα,v is given by

ψ (˜e) := # e (v,αv) e e˜ (mod Γ) # e (v,αv) e e˜ (mod Γ) α,v { ∈ | ≡ } − { ∈ | − ≡ }

and this map is independent of the vertex v, where (v,αv) denotes the path without backtracking from v to αv.

For a proof of this lemma see [Non01, Thm. 2.34]. Here we give a short explanation of the meaning of the map ψ . Given any α Γ, the path on the tree T that goes from v to α,v ∈ αv is non back-tracking. However, its projection to the quotient graph is a cycle which in general is not reduced. Given anye ˜ in the cycle determined by v and αv, there are many edges e T mapping toe ˜. So the map ψ counts how many edges of the path v,αv in T ∈ α,v have the image in the cycle.

Lemma 2.9.4. There exists an injective homomorphism

ι : H (Γ T, Z) H (T, Z)Γ 1 \ → ! ψ ϕ 7−→ 1 deﬁned by ϕ(e) := n(e)ψ(˜e), where n(e) := #Z(Γ)− #StabΓ(e).

22 2.9. Harmonic cocycles

¯ Γ Finally, the following lemma connects the groups Γ and H!(T, Z) .

Lemma 2.9.5. There is a canonical homomorphism

j : Γ¯ H (T, Z)Γ → ! α ϕ 7−→ α,v

1 where ϕα(e) := ϕα,v(e)= #(Z Γ) γ Γ δ(e,α,v,γ) with v any ﬁxed vertex and the function ∩ ∈ δ(e,α,v,γ) given by: P

1 if γ(e) (v,αv), ∈ δ(e,α,γ,v)=  1 if γ(e) (αv,v),  − ∈  0 otherwise.   This homomorphism is independent of the choice of the vertex v.

The map j deﬁned above is actually the composition of i and ι, so it is injective. We show here the surjectivity of j (cf. [Non01, Cor. 2.37]) which will give us a way to construct the space of harmonic cocycles. Here we will use the homology of the graph to construct Γ a basis of H!(T, Z) and then we get the surjectivity.

1 Let T be the maximal tree in (Γ T)◦ and e˜ , e˜ , ..., e˜ be a set of representatives of the \ { 1 2 g} edges of the tree (Γ T)◦ T . It is proved in [Non01, Cor. 2.38] that the set e˜ , e˜ , ..., e˜ \ − { 1 2 g} consist of edges attached to vertices of level 0 of degree q + 1 (cf. .2.8 for the deﬁnition of § level).

Letv ˜i = o(˜ei),w ˜i = t(˜ei) the origin and target ofe ˜i, respectively. Then there exists a ′ ′ unique geodesicc ˜i in the tree T joiningw ˜i andv ˜i. In this wayc ˜i gives rise to a closed path ′ c in Γ T by composinge ˜ andc ˜ . i \ i i Consider ϕ : Y (Γ T) T i \ −→ deﬁned as follows

1It exists since (Γ T)◦ is ﬁnite and connected. \

23 2. Background

n(e) ife ˜ appears inc ˜i,

ϕi(˜e)= n(e) if e˜¯ appears inc ˜ , − i  0 otherwise.

 Γ From (2.5), we see that actually ϕi lifts to a function on H!(T, Z) and form a basis of H (T, Z)Γ. Finally let c = (e , e , ...e ) be a lift ofc ˜ , then there is a α Γ such that ! i i,0 i,1 i,l i i ∈ α o(e )= t(e ) so we can find a α Γ such that j(α) H (T, Z)Γ. i i,0 i,l ∈ ∈ ! In summary, to construct the space of harmonic cocycles for the graph Γ T, we need to \ find the maximal subtree T of (Γ T)◦, also called maximal spanning tree. For this we \ use the algorithm given in [Non01, 3] to find a set of representatives e˜ , e˜ , ..., e˜ of the § { 1 2 g} edges of the tree (Γ T)◦ T . Finally we define for each i the ϕ as above. \ − i

2.10 Theta functions for arithmetic groups

Let Γ be an arithmetic subgroup of GL2(K ). ∞

Deﬁnition 2.10.1. A holomorphic theta function for Γ GL2(K ) is an invertible rigid ⊂ ∞ analytic function (u : Ω C× ) OΩ(Ω)× such that for each γ Γ, u satisﬁes the −→ ∞ ∈ ∈ functional equation

u(γz)= cu(γ)u(z) for some constant cu(γ) C× independently of z. ∈ ∞

The map cu : γ cu(γ) is a homomorphism from Γ to C× called the multiplier of u. 7→ ∞

We denote by Θh(Γ) the space of holomorphic theta functions for Γ.

In [GR96, 5], the authors give a way to construct a holomorphic theta function for Γ as § follows. Let ω be ﬁxed element of Ω and α Γ, put ∈ z γω θ (ω, z) := − . (2.6) α z αγω γ Γ˜ − Y∈ Theorem 2.10.2 ([GR96, Thm. 5.4.1]). The product θα(ω, z) converges to a holomorphic theta function for Γ on Ω. The value of c (γ) := θα(ω,γz) induces a group homomorphism α θα(ω,z) c¯ : Γ¯ Hom(Γ¯,C× ) by α cα. Moreover, the map Γ¯ Γ¯ C× deﬁned by (α, β) −→ ∞ 7→ × −→ ∞ 7→ cα(β) is a symmetric bilinear pairing.

24 2.10. Theta functions for arithmetic groups

Corollary 2.10.3 ([GR96, Cor. 5.4.12]). The constant cα(β) lies in K× C× . ∞ ⊂ ∞

In the next chapter we will see another way to construct theta functions by means of certain multiplicative integrals.

2.10.1 Theta functions and harmonic cocycles

In this subsection we relate the theta function for an arithmetic group Γ and the Z-valued harmonic cocycles by means of the following result due (mostly) to Van der Put [VdP82].

Theorem 2.10.4. The map

f t(e) r : O (Ω)× H (T, Z), given by r(f)(e) := log k k Ω −→ ! q f k ko(e) is a continuous surjective group homomorphism with kernel C× where v is the spectral ∞ 1 k·k norm on OΩ(λ− (v)) deﬁned by

1 f := sup f(z) z λ− (v) k kv | || ∈ for v X(T). ∈

The next result is a reﬁnement of the previous one, since it relates theta functions to Γ-invariant harmonic cocycles (cf. [GR96, Thm. 5.6.1]).

Theorem 2.10.5. Let α Γ be given then r(θ )= ϕ . ∈ α α

The map r yields a map Γ r¯ :Θh(Γ)/C× H!(T, Z) . ∞ −→ Let further

u¯ : Γ¯ Θh(Γ)/C× −→ ∞ be the map induced from α θ , then we have the following: 7→ α Theorem 2.10.6 ([GR96, 6]). For α Γ with class α¯ in Γ¯ we have that r¯(θ ) = j(¯α). § ∈ α In other words, the following diagram is commutative ¯ Γ ❙❙ ❙❙❙❙ ❙❙❙❙ j u¯ ❙❙❙❙ (2.7) ❙❙❙❙) r¯ / Γ Θh(Γ)/C× H!(T, Z) . ∞

25 2. Background

From the theorem we see that we have two diﬀerent constructions of Γ-invariant Z-valued harmonic cocycles from α Γ: ∈

i) using the function ϕα given in Lemma 2.9.5 and

ii) by evaluating the mapr ¯ in θα.

From now by abuse on notation, we will write r instead ofr ¯.

2.11 Hecke operators

There exist Hecke operators acting on each of the groups that appear in (2.7). For the

deﬁnition of the operators on Γ¯ and Θ(Γ)/C× see [GR96, 9.3], we will give an explicit ∞ § Γ description of the operator on H!(T, Z) for the case Γ = Γ0(N).

Let m be a non zero ideal of Fq[T ]. We recall from the deﬁnition of the Bruhat-Tits tree that the edges are given by classes of GL2(K )/K× I. Hence functions on Y (T) can be ∞ ∞ seen as functions on GL2(K ) right invariant under I. For ϕ on GL2(K ) we put ∞ ∞

Tmϕ(α) := ϕ(γα) γ Rm X∈ where

a b Rm = GL (A) a, d monic (ad)= m, gcd(a, N)=1, deg b< deg d . c d ∈ 2 (2.8)

The following properties of Tm are standard:

Γ i) Tm : ϕ Tmϕ maps H (T, Z) into itself, 7→ !

ii) all Tm commute,

iii) if m and n are coprime, then Tmn = Tm Tn, ◦

deg p iv) if gcd(p, N) = 1 then T n+1 = Tpn Tp q T n−1 for p a prime ideal of A, p ◦ − p

26 2.12. Application to the Shimura-Taniyama-Weil uniformization

v) if gcd(m, N) = 1 then Tm is hermitian with respect to the Peterson inner product deﬁned as follows:

(ϕ1,ϕ2)(e)= ϕ1(e)ϕ2(e). e Γ T X∈ \

Hecke operators Tm with gcd(m, N) = 1 are called unramiﬁed. As in the classical case we can also construct the space of new and old forms. Indeed suppose that M divides N. For each monic divisor a of N/M we have an embedding i : H (T, Z)Γ0(M) H (T, Z)Γ0(N) a,M ! −→ ! given by a 0 i (ϕ)(g)= ϕ( g). a,M 0 1 We set then (H (T, Z)Γ0(N) Q)new to be the orthogonal complement in H (T, Z)Γ0(N) Q ! ⊗ ! ⊗ with respect to the Perterson inner product to the images of all the i Q, where M a,M ⊗ runs through the proper divisors of N and through the divisors of N/M. We further put

Hnew(T, Z)Γ0(N) = H (T, Z)Γ0(N) (H (T, Z)Γ0(N) Q)new. ! ! ∩ ! ⊗

2.12 Application to the Shimura-Taniyama-Weil uniformization

In order to establish the analog of the classical Shimura-Taniyama-Weil, we need to explain how Q-harmonic cocycles may be regarded as automorphic forms of a certain type. For a deeper discussion of automorphic forms we refer the reader to [Gel75].

In what follows, A will be the adèle ring of K with ring of integers O and I the idèle group. Having fixed the place of K, these rings decompose into a “finite” and an “infinite” ∞ part A = Af K × ∞ O = Of O × ∞ I = If K× . × ∞

We deﬁne an automorphic cusp form for an open subgroup K of GL2(O)tobea C-valued

function ϕ on Y (K) := GL2(K) GL2(A)/K Z(K ). \ · ∞

Let S be a set of representatives for GL (K) GL (A )/K and deﬁne for x S 2 \ 2 f f ∈ 1 Γ := GL (K) xK x− . x 2 ∩ f

27 2. Background

In [GR96, 4] the following isomorphism is shown: §

∼= GL2(K) GL2(A)/Kf Z(K ) I x S Γx GL2(K )/Z(K ) I \ × ∞ · −→ ∈ \ ∞ ∞ · F = x S X(Γx T). ∈ \ We have the following theorem. F

Theorem 2.12.1 (Drinfeld). Let K be an open subgroup of GL2(O) of the form K = (Kf) I and let F be a ﬁeld of characteristic zero. Under the previous bijection the module × of harmonic cochains

Γx H!(T, F ) x S M∈ corresponds to the space Wsp(K, F ) of F -valued cuspidal automorphic forms ϕ on

G(K) GL2(A)/Kf Z(K ) I that transform like πsp under GL2(K ). \ × ∞ · ∞

Here πsp is the so-called special representation of GL2(K ), i.e., the irreducible represen- ∞ 1 tation of GL2(K ) on the space of locally constant F -valued functions on P (K ). For ∞ ∞ more details see [GR96, 4.7]. § So the space of harmonic cocycles has a natural interpretation as a space of automorphic functions in the sense of Jacquet-Langlands [JL70].

Let E be an elliptic curve deﬁned over K = F (T ) with conductor N , where N is an q ∞ ideal of A and assume E to have split multiplicative reduction at . By combining results ∞ of Jacquet-Langlands [JL70], E corresponds to an automorphic eigenform ϕE in the new

Γ0(N) part of H!(T, Z) with rational integral eigenvalues. The elliptic curve E is an isogeny

factor of the new part of the Jacobian of the Drinfeld modular curve X0(N). This is the content of [GR96, 8.3]. § Applying well known facts from the theory of automorphic forms there are canonical bijections between the following three sets

i) K-isogeny classes of elliptic curves with conductor N , ∞

new ii) one-dimensional isogeny factors of J0 (N),

new Γ0(N) iii) normalized Hecke eigenforms ϕ in H! (T, Z) with rational eigenvalues.

28 2.12. Application to the Shimura-Taniyama-Weil uniformization

Thus one would like to have a procedure to construct an elliptic curve within the isogeny class that corresponds to the newform ϕ.

Let ϕ Hnew(T,G)Γ0(N) be a Hecke eigenform with rational eigenvalues. We will construct ∈ ! Γ0(N) ¯ the elliptic curve Eϕ associated to it. By means of (2.7) we identify H!(T,G) with Γ.

Proposition 2.12.2 ([Gek95, Thm. 3.2]). Let ϕ j(Γ) be an harmonic cocycle, regarded ∈ as the class of some element, also denoted by ϕ, of Γ. Put ∆ K× for the subgroup ⊂ ∞ cϕ(α) α Γ of K× , where cϕ(α) is the multiplier associated to the theta function of Γ. ∞ { | ∈ } Z Then there exists q K× such that q < 1 and q = ∆. ∈ ∞ | |

For general A, the conclusion of Proposition 2.12.2 is that there exists q K× with q < 1 ∈ ∞ | | such that ∆ qZ and the incluision is of ﬁnite index (cf. [GR96, Prop 9.5.1]). ⊇

Then the elliptic curve Eϕ associated with an automorphic Hecke eigenform, can be ana- an Z lytically recovered as a Tate curve (cf. Ch. 5, 5.5 for a definition) as Eϕ (C )= C× /q . § ∞ ∞ It is shown in [Roq70, 3] that E may be described by means of an analytic equation § ϕ defined over a finite extension of K. However, this is not enough to recognize the isogeny class of the elliptic curve Eϕ.

The aim of this thesis is calculate by means of a certain multiplicative integral the Tate period q and then using the analytic equation of the Tate curve and an appropriate model of an elliptic curve that by means of change of variables allows us to ﬁnd the algebraic equation of the elliptic curve Eϕ over K.

29 30 3. Integration, Theta function and uniformizations

3.1 Integration

As described in the previous chapter, Gekeler and Reversat develop in [GR96] the theory of theta functions to construct an explicit analytic parametrization of an elliptic curve with semi-stable reduction at the place . However, their construction of the theta function ∞ requires to compute the infinite product (2.6), which makes it computationally hard to find the Tate parameter. In [Lon02], Longhi defines a multiplicative integral over P1(K ) ∞ which is a multiplicative version of Teiltebaum’s Poisson formula [Tei91], and uses this to construct a theta function in a different way as the one of Gekeler-Reversat. Around the same time Pal gives a similar construction in [Pál06]. In this section we briefly recall the main facts of the machinery of integration along the lines of Longhi [Lon02].

Let X be a topological space such that its compact open subsets form a basis for the topology.

Deﬁnition 3.1.1. A Z-valued function µ on compact-open subsets of X is a distribution on X if µ(X) = 0 and if it is ﬁnitely additive, i.e., if, whenever A and B are disjoint compact open sets of X, one has µ(A B) = µ(A) + µ(B). We denote the space of ∪ Z-valued distributions by M(X, Z).

Deﬁnition 3.1.2. A measure on X is a bounded distribution, i.e., a distribution µ for which there is a constant C satisfying µ(U)

31 3. Integration, Theta function and uniformizations

From now on, we assume that X is compact, so in particular Z-valued distributions on X are measures.

Deﬁnition 3.1.3. Given a continuous function f : X C× , its multiplicative integral → ∞ with respect to the measure µ M (X, Z) is ∈ 0

f(t) dµ(t) := lim f(u)µ(U) (3.1) ×X −→α U Cα Z Y∈ where C is the direct system of ﬁnite covers of X by compact open subsets U and u { α}α is an arbitrary point in U.

Under the crucial assumption that X is compact, we have

Proposition 3.1.4 ([Lon02, Prop. 5]). The limit in (3.1) exists and is independent of the choice of the u’s. Furthermore

dµ : C(X,C× ) C× (3.2) × ∞ → ∞ Z X

is a continuous homomorphism, where C(X,C× ) is the set of continuous functions from ∞ C× to X. ∞

3.1.1 Measures and harmonic cocycles

We know from Lemma 2.5.4 that the set of ends of the tree T is in 1-1 correspondence with P1(K ). ∞ Proposition 3.1.5. The map

M (X, Z) H(T, Z) 0 −→ µ (e µ(U )) 7−→ 7→ e is an isomorphism of Z-modules.

1 Proof. Let e be an oriented edge of the tree T and let the open subset Ue of P (K ) be ∞ defined as in Section 2.5. For every open compact subset U of P1(K ) there is a finite set ˙ ∞ YU of oriented edges such that U = e Y Ue. So, a ϕ in H(T, Z) defines a finite additive ∈ U measure µϕ M0(X, Z) by putting µϕ(U)= e Y ϕ(e). ∈ S ∈ P

32 3.1. Integration

Conversely given a Z-valued measure µ M (X, Z) one deﬁnes the harmonic cocycle ∈ 0 ˙ 1 1 ϕµ(e) := µ(Ue). From Ue Ue = P (K ) and µ(P (K )) = 0 it follows that ϕ(e)= ϕ(e). ∞ ∞ 1 − Let now v be a vertex in T, then P (K ) is the disjoint union of Ue’s, where the union is S ∞ taken over the oriented edges starting at v. Again from the fact that µ(P1(K )) = 0 and ∞ µ(U V )= µ(U)+ µ(V ) if U V = we get ϕ (e) = 0. We conclude that ϕ is indeed ∪ ∩ ∅ e µ µ a harmonic cocycle. P

3.1.2 The integral over ∂Ω

We are interested in the particular case where X = ∂Ω. In this case the computation of a multiplicative integral can be accomplished as follows. Choose a vertex v X(T). Usually ∈ the vertex v is taken to be v as we will do, and for each e Y (T) pointing away from v , 0 ∈ 0 deﬁne l(e) to be the distance between the origin o(e) of e and v0. Then according to the deﬁnition of the integral (3.1) and the discussion in Subsection 3.1.1 we have that

µ(Ue) f(t) dµϕ(t) = lim f(te) (3.3) n ×∂Ω →∞ Z l(Ye)=n

is independent of the choice of v0.

3.1.3 Change of variables

1 Note that the group GL2(K ) acts naturally on the space M0(P (K ), Z) by γ µ(U) := ∞ ∞ ∗ 1 1 1 µ(γ− U) for any µ M0(P (K ), Z) and U an open in P (K ). In the same way GL2(K ) ∈ ∞ ∞ ∞ 1 acts on the space of harmonic cocycles H(T, Z)by(γ ϕ)(e) := ϕ(γ− e). As in the classical ∗ case, we have the formula of change of variables formula given by

f d(γ µ)= (f γ) dµ × ∗ × ◦ Z γU Z U

or equivalently

1 1 (γ− f) d(γ− µ)= f dµ (3.4) × −1 ∗ ∗ × Z γ U Z U 1 where (γ− f)(t)=(f γ)(t)= f(γ t ). ∗ ◦ h i

33 3. Integration, Theta function and uniformizations

3.2 Theta function

As already observed in [Sch84], the machinery of integration on ∂Ω can be used to construct an inverse of the Van der Put’s map r deﬁned in the equation (2.10.4). In [Lon02], Longhi uses the multiplicative integral to give a multiplicative version of Teitelbaums’s Poisson formula [Tei91, Thm. 11] and then such an inverse of r. In the following paragraphs we will describe the Poison formula stated by Longhi and then the integral that we are interested in.

Theorem 3.2.1 ([Lon02, Thm. 6]). Let u O (Ω), ϕ = r(u) and ﬁx z Ω. Then for ∈ Ω 0 ∈ z Ω ∈ z t u(z)= u(z ) − dµ (t). (3.5) 0 × z t ϕ Z∂Ω 0 −

We are now ready to prove

Γ Proposition 3.2.2. The map r induces an isomorphism H!(T, Z) = Θh(Γ)/C . ∼ ∞

Proof. Given a u Θh(Γ)/C then by Theorem 2.10.4 we have that r(u) is a harmonic ∈ ∞ cocycle, Γ-invariant and cuspidal.

Conversely given a harmonic cocycle ϕ H (T, Z)Γ we denote by µ its corresponding ∈ ! ϕ measure. Fix a z0 in Ω. Consider now the function

z t u(z)= − dµ (t). (3.6) × z t ϕ Z∂Ω 0 −

By Theorem 3.2.1 we know that u O (Ω)× . It remains to check that u(z) satisﬁes the ∈ Ω functional equation

u(γz)= cu(γ)u(z) (3.7)

for all γ Γ0(N) and some constant cu(γ) C× . ∈ ∈ ∞

a b It is a straightforward calculation to see that for γ =( c d ) , the ratio

t γz − t γz0 cz + d − γ−1t z = (3.8) −1 − cz0 + d γ t z0 −

34 3.3. Complex uniformization

does not depend on t. So we have t γz u(γz)= − dµ (t) × t z ϕ Z∂Ω − 0 t γz0 t γz = − dµϕ(t) − dµϕ(t) ×∂Ω t z0 ×∂Ω t γz0 Z − Z − 1 t γz cz + d γ− t z = − 0 dµ (t) − dµ (t). × t z ϕ × cz + d γ 1t z ϕ Z∂Ω − 0 Z∂Ω 0 − − 0 Using the fact that the integral is multiplicative and µϕ(∂Ω) = 0 we get that cz + d 0 dµ (t)=1, × cz + d ϕ Z∂Ω since it is constant on t. The functional equation follows taking t γz c (γ)= − 0 dµ (t) (3.9) u × t z ϕ Z∂Ω − 0 which does not depends on z.

We recall the diagram (2.7) from Chapter 2. ¯ Γ ❙❙ ❙❙❙❙ ❙❙❙❙ j u¯ ❙❙❙❙ ❙❙❙❙) r¯ / Γ Θh(Γ)/C× H!(T, Z) . ∞ So given a harmonic cocycle ϕ H (T, Z)Γ and α Γ¯ so that j(α)= ϕ, we have that the ∈ ! ∈ multiplier cu(γ) is actually cα(γ) deﬁned in Theorem 2.10.2.

3.3 Complex uniformization

In this section we consider an elliptic curve E/Q with conductor N. We will brieﬂy describe how to parametrize the curve E over the complex ﬁeld C. The main reference here is [Sil09, Ch. VI].

Let Γ0(N) be the group of matrices in SL2(Z) which are upper triangular modulo N. It acts as a discrete group by M¨obius transformations on the Poincare upper half-plane

H := z C Im(z) > 0 . { ∈ | }

35 3. Integration, Theta function and uniformizations

A cusp form of weight 2 for Γ0(N) is an analytic function f on H satisfying the relation a b f(γ z )=(cz + d)2f(z) for all γ = Γ (N), (3.10) h i c d ∈ 0 together with suitable growth conditions on the boundary of H. The invariance equation (3.10) implies in particular that the function f is periodic of period 1 and thus f can be written as a power series in q = e2πi with no constant term:

∞ n f(z)= anq . n=1 X s The associated L-series is deﬁned as L(f,s) := n∞=1 ann− .

From the Eichler-Shimura theory given a cuspidalP eigenform whose Fourier coeﬃcients are

integers, there exists an elliptic curve Ef such that the two L-series coincide

L(f,s)= L(Ef ,s).

Here L(Ef ,s) is the L-series deﬁned as the inﬁnite Euler product

s 1 2s 1 s 1 s L(E ,s)= 1 a p− + p − − (1 a p− )− := a n− f − p − p n p∤N p N Y Y| X with a = #E(F ) ([Dar04, 1.4]). p p § On the other hand, let E be an elliptic curve deﬁned over Q of conductor N. Then there

exists a cuspidal Hecke eigenform of weight 2 for Γ0(N) such that the L functions coincide

L(f,s)= L(E,s)

and E is isogenous to the elliptic curve Ef obtained form Eichler-Shimura theory.

In summary, given an integer N > 1 and a cuspidal Hecke eigenform of weight 2, one would like to have a procedure to construct an elliptic curve within the isogeny class that correspond to the form f. Actually such as procedure exists and we will brieﬂy give some things related to it (a good reference here is [Cre97]).

Let X0(N) be the modular elliptic curve for cyclic N-isogenies. By the work of Wiles et.al., E is equipped with a non constant dominant morphism deﬁned over Q, commonly referred as the Weil parametrization attached to E:

Φ : X (N) E N 0 −→

36 3.3. Complex uniformization

mapping the cusp to the identity element of E. The complex uniformization of E(C) ∞ provides a method for calculating the Weil parametrization. Namely, the compact Riemann surface E(C) is isomorphic to C/Λ, where Λ is a lattice generated by the periods of a N´eron diﬀerential ω on E. Then we have the following commutative diagram

z0 z0 R f (z)dz 7→ ∞ E / Γ H∗ C/Λ (3.11) \ j η ΦN / X0(N)(C) E(C)

where η : C/Λ E(C) is the complex analytic isomorphism described by the formula −→ η =(℘Λ : ℘Λ′ : 1) where ℘Λ is the Weierstrass ℘-function attached to Λ. Explicit formulas

for Λ and ℘Λ can be found in [Sil09].

τ In this case a good approximation of the integral fE(z)dz reduces to calculate a ﬁnite ∞ sum which depends on the calculation of the an’s inR the Fourier expansion of fE where fE is the modular form attached to E.

Obtaining equations for the curves

z0 The integrals fE(z)dz allow to compute periods ω1 and ω2 which generate the period ∞ lattice Λf of theR modular curve Ef = C/Λf . Letting τ = ω1/ω2, we may assume that 2πiτ Im(τ) > 0 interchanging ω1 and ω2 if necessary. Set q = e and deﬁne

4 2π ∞ n3qn c (q)= 1+240 and 4 ω 1 qn 2 n=1 − ! 6 X 2π ∞ n5qn c (q)= 1 504 . 6 ω − 1 qn 2 n=1 ! X −

Then the following theorem is crucial in the calculation of the equation of the curve E (cf. [Cre97, 2.14]). §

Theorem 3.3.1 (Edixhoven). The quantities c4 and c6 deﬁned above are in Z, so the elliptic curve y2 =4x3 c x c − 4 − 6 is deﬁned over Z.

37 3. Integration, Theta function and uniformizations

This theorem allows Cremona to ﬁnd equations for all elliptic curves of conductor up to

130.000 [Cre97]. The series c4 and c6 converge extremely rapidly. Thus, assuming that

ω1 and ω2 are known to suﬃcient precision, Cremona can compute c4 and c6 also with suﬃcient precision and thus he is able to recognize the corresponding exact integer values.

3.4 p-adic Uniformization

From the previous section, we have that any elliptic curve E/Q with conductor N is equipped with a non constant parametrization

Φ : X (N) E. (3.12) N 0 −→ However for some applications, like the calculation of Heegner points, (cf. [Dar04, Ch. 3-4]) it is convenient to enlarge the repertoire of modular parametrizations to include Shimura curve parametrizations. For more details on the results stated here, the reader is referred to [Dar04],[BC91],[Voi] and [Vig80].

+ Assume that the positive integer N is square free and N = N −N is a factorization of N

such that N − has an even number of prime factors. Let C be the indeﬁnite quaternion

Q-algebra ramiﬁed precisely at the primes dividing N − and let S be an Eichler Z-order in C of level N +. Fix an identiﬁcation

ι : C Q R = M2(R). ∞ ⊗ ∼

Denote by ΓN −,N + the image under ι of the group of units in S of reduced norm 1. ∞ Then ΓN −,N + acts properly discontinuously on H with compact quotient XN −,N + (C). By

Shimura theory, the compact Riemann surface XN −N + (C) has a canonical model XN −,N +

over Q as in the classical case. This is done by interpreting XN −,N + as a moduli space for abelian surfaces over Q with endomorphism rings containing S, together with some auxiliary level N +-structure (cf. [AB04]).

Let JN −,N − denote the Jacobian of XN −,N + . By the modularity theorem for elliptic curves deﬁned over Q and the Jacquet-Langlands correspondence, there exists a surjective morphism

Φ − + : J − + E (3.13) N ,N N ,N −→ deﬁned over Q (cf. [Dar04, Ch. 4]).

38 3.4. p-adic Uniformization

However, we do not dispose here of Fourier coeﬃcients, since modular forms on non-split quaternion algebras do not admit q-expansions, there is no known explicit formula for the

map ΦN −,N + . In order to handle with this issue, it is necessary to turn to the p-adic uniformization. In this section we will succinctly explain how to use the uniformization

ΦN −,N + to construct an explicit p-adic uniformization of E at the primes p dividing N −.

Let us assume N − > 1 and p be a prime dividing N −. Consider now the definite quaternion algebra B ramified precisely at the infinity place together with the primes dividing N −/p and let R be an Eichler Z-order in B of level pN +. Choose an identification

ι : B Q M (Q ). p ⊗Q p −→ 2 p (p) Let ΓN −,N + GL2(Qp) be the image under ιp of the group of units in R of reduced norm ⊂ (p) 1. In the remaining part of this section we will write Γ instead of ΓN −,N + .

+ Let p be a ﬁxed prime and let N be a square-free integer that factorizes as pN −N such that N − has an odd number of prime factors. Such a factorization is called an admissible factorization.

We recall here some objects already defined in the function fields case, like the upper half plane, the Bruhat Tits tree, distributions, measures, etc. These concepts are necessary to define our p-adic integral.

Let Cp be a p-adic completion of an algebraic closure of Qp. As in Chapter 2, we deﬁne the p-adic upper half plane to be

H := P1(C ) P1(Q ) p p − p with the action of GL2(Qp) by linear fractional transformations. Similarly as in Chapter

2, one has deﬁnitions of admissible covering by sets An and the space of rigid analytic functions on Hp.

The group Γ acts on Hp with compact quotient Γ Hp, which is equipped with the structure \ (p) of a rigid analytic curve over Qp. It can be identiﬁed with an algebraic curve X over ΓN−,N+ Qp (cf. [GvdP80]).

The Bruhat Tits tree Tp of PGL2(Qp) is a connected p + 1-regular tree with set of vertices 2 X(Tp) deﬁned as classes of homothety Zp-lattices in Qp. The set of edges Y (Tp) consists of pair of vertices (v0, v1) represented by lattices λ1 and Λ2 such that pΛ2 Λ1 Λ2.

39 3. Integration, Theta function and uniformizations

The corresponding identiﬁcations given by Propositions 2.4.1 and 2.4.3 are still valid for vertices and edges, respectively.

Let Γ be as above. A Γ-invariant harmonic cocycle with values in an abelian group B is a map ϕ : X(T ) B such that p −→

1) ϕ(¯e)= ϕ(e) for all e Y (T ), − ∈ p 2) For all v X(T ) we have ϕ(e)=0, ∈ p t(e)=v 3) ϕ(γe)= ϕ(e) for all γ Γ.P ∈

Deﬁnition 3.4.1. Let f be a rigid analytic function of Hp with values in Cp, we say that f is a rigid analytic modular form of weigh k on Γ H if \ p f(γ τ )=(cτ + d)kf(τ) h i for all γ =( a b ) Γ and τ H . c d ∈ ∈ p

The Cp-vector space of rigid analytic modular forms of weight k with respect to Γ is denoted

by Sk(Γ).

1 Deﬁnition 3.4.2. (Distribution and measure) A p-adic distribution on P (Qp) is a ﬁnitely 1 1 addictive Cp-valued function µ on the compact open sets of P (Qp) satisfying µ(P (Qp)) = 0. If µ is p-adically bounded then it is called a p-adic measure. The space of all measures 1 1 on P (Qp) is denoted by Meas(P (Qp), Cp).

As in the case of function fields, we have an action of GL2(Qp) on the space of measures and on the space of harmonic cocycles (cf. 3.1.3 for the definition) and the corresponding § 1 identification between the space of measures on P (Qp) and the space of harmonic cocycles.

1 Let f be any continuous function on P (Qp) then we deﬁne the integral of f with respect 1 to a measure µ on P (Qp) as

f(t) dµ(t) = lim f(u)µ(U) 1 P (Qp) α −→ U Cα Z X∈ where the limit is taken over increasing ﬁne covers C of P1(Q ) by disjoint open { α}α p compact subsets U.

40 3.4. p-adic Uniformization

1 Γ 1 Denote by Meas(P (Qp), Cp) the space of all Γ-invariant measures on P (Qp). There is a well known theorem due to Schneider and Teitelbaum that gives the following isomorphism (cf. [Dar04, Thm.5.9]) 1 Γ Meas(P (Qp), Cp) ∼= S2(Γ). We have that Meas(P1(Q ), Z)Γ Meas(P1(Q ), C )Γ given by the Z-valued harmonic p ⊂ p p cocycles. It thus, gives rise via the Scheneider-Teitelbaum isomorphism to an integral structure S (Γ)Z S (Γ), it plays a role somewhat similar to that of modular forms with 2 ⊂ 2 integral Fourier coeﬃcients in the theory of classical modular forms.

1 Fix an extension log : C× C of the p-adic logarithm to all C×. p p −→

Deﬁnition 3.4.3. Let f be a rigid analytic modular form of weight two for Γ and µf be the measure attached to it by the Scheneider-Teitelbaum isomorphism. Fix τ , τ H . 1 2 ∈ p The p-adic line integral attached to f(z)dz is deﬁned to be

τ2 t τ2 f(z) dz := logp − dµf (t). (3.14) 1 t τ Zτ1 ZP (Qp) − 1

If we formally exponentiate the expression (3.14) we get the multiplicative line integral

τ2 t τ2 f(z) d(z)= − dµf (t). (3.15) × × 1 t τ Z τ1 Z P (Qp) − 1 This multiplicative integral is more canonical than its additive counterpart since it does not depend on a choice of a branch of the p-adic logarithm. In fact one should deﬁne the right hand side of (3.15) as in (3.1).

Let E be an elliptic curve over Q of conductor N, which admits an admissible factorization + as N −N with p dividing N − and let Γ SL2(Qp) be the discrete subgroup arising from ⊂ (p) this factorization as explained above. The rigid variety X and the curve XΓ − + ΓN−,N+ N ,N are connected by the following theorem due to Cerednik and Drinfeld.

Theorem 3.4.4. There is a canonical rigid analytic isomorphism

(p) CD : XΓ − (Cp) XΓ − + (Cp). N ,N+ −→ N ,N

1 By imposing logp(wz) = logp(w) + logp(z) and setting logp(p) = 0.

41 3. Integration, Theta function and uniformizations

The Cerednik-Drinfeld theorem together with the following isomorphisms ([GvdP80, Ch. VI and VIII])

0 Div (Γ − + H) = J − + N ,N \ ∼ N ,N 0 (p) (p) Div (Γ − + H ) = J − + , N ,N \ p ∼ N ,N 0 (p) gives rise to a map Div (Γ − H ) J − + (C ) also denoted CD by abuse of N ,N + \ p −→ N ,N p notation.

On the other hand, we can use the multiplicative integral (3.14) to deﬁne a p-adic Abel- Jacobi map

0 (p) Z Φ : Div (Γ − H ) Hom(S (Γ) , C×) C× AJ N ,N + \ p −→ 2 p ≃ p τ2 τ2 τ1 (f f(z) dz)). − 7−→ 7→ ×τ1 R Let Φ : C× E (C ) be the Tate parametrization of E (cf. [Sil94]). Assume that the Tate p −→ q p curve E is a strong Weil curve, if not replace it by an isogenous curve. Then we have the following result

Proposition 3.4.5. The following diagram commutes

0 (p) ΦAJ / Div (Γ − + H ) C× N ,N \ p p CD ΦTate ΦN / JN −,N + (Cp) E(Cp).

For a discussion of this result, see [BD98]. Compare with the diagram (3.11).

Following the ideas of Pollack and Stevens [PS11], Greenberg [Gre06] is able to give an algorithm, running in polynomial time, for evaluating the p-adic integral (3.15). In the remaining of this chapter, we will brieﬂy describe the method devised by Greenberg.

Let A := v(x)= a xn a Q , a 0 as n . (3.16) rig n n ∈ q n → →∞ ( n>0 ) X

Elements of Arig are rigid analytic functions on the closed unit disk in Cp which are deﬁned over Qp.

Deﬁnition 3.4.6. A rigid analytic distribution µ is an element of the continuous dual of

Arig. The space of rigid analytic distributions is denoted by Drig.

42 3.4. p-adic Uniformization

There is no problem in deﬁning the distribution in this way since it “restricts” to a distribution as one of Deﬁnition 3.4.2 (cf. [Kob84, Ch. II, 3]). §

Let µ Drig, v Arig and consider the integral v(x) dµ(x). According to [PS11] the ∈ ∈ Zp problem to compute it reduces to the calculationR of the moments µ(xn)= xn dµ(x) n > 0. ZZp M+1 n Then we say that µ is known to precision M > 1 if v(x) dµ(x) is known modulo p − Zp for n =0, ..., M. R Deﬁnition 3.4.7. Amap Φ:Γ GL (Q ) D that is I -equivariant (for Iwahori I \ 2 p → rig p p subgroup) under the right action is called a overconvergent modular form for Γ. The space

of overconvergent modular forms for Γ is denoted by M2(Γ).

f In [PS11] Pollack and Stevens deﬁne a Hecke operator Up on the space M2(Γ). It is 2 also proved there that its Up -invariant subspace is Hecke-isomorphic to the space of rigid f analytic modular forms for Γ. Then, given a rigid analytic modular form f with rational

Hecke eigenvalues, one can ﬁnd its corresponding Z-valued harmonic cocycle ϕf and the measure µ with U f = f (cf. [Gre06, 5]). In order to calculate the moments attached f p ± § to the distribution µf it is enough to calculate the values of the corresponding element Φ M (Γ) which can be computed up to precision M from ϕ (mod p) by iterating the f ∈ 2 f operator Up in time O(M). f We come back now to the calculation of the p-adic line integral. Denote by

t τ2 J(τ2, τ1)= − dµf . × 1 t τ Z P (Qp) − 1 Under some technical hypothesis, it is enough to compute log J(τ2, τ1). Write

t τ2 log J(τ2, τ1)= log Ja(τ2, τ1) where Ja(τ2, τ1)= − dµf (3.17) 1 ×ba t τ1 a P (Fp) Z ∈X − and b is the standard residue disk around a (cf. [Gre06, 8]) deﬁned as a § b := x Q x a < 1/p . a { ∈ p | | − | }

1 Note that ba = Ue after identifying P (Fp) with the edges with target v0.

These p + 1 integrals can be calculated in polynomial time by evaluating certain moments coming from the expansion of the function logarithm as a series (cf. [Gre06, Prop. 14]).

43 3. Integration, Theta function and uniformizations

Remark 3.4.8. The use of the logarithm to calculate the p-adic multiplicative line integral is crucial. We do not dispose of this function in the function ﬁeld case and this is one of the main diﬃculties we have to overcome.

+ Let N be an integer which admits a factorization N −N with a prime p dividing N − and (p) let f be a rigid analytic modular form for ΓN −,N + with rational Hecke eigenvalues. Deﬁne

t γτ1 the set Λf = 1 − dµf γ Γ , which is discrete in C×. Set qf as the generator ×P (Qp) t τ1 ∈ p − of Λf . Then qf Ris the Tate parameter attached to the form f.

As in the complex case one would like to have a procedure to find the elliptic curve Ef with coefficients in Q. Even though Greenberg managed to calculate with good accuracy qf , it is not clear whether one can compute from it an equation for the elliptic curve Ef with coefficients in Q as Cremona does.

44 4. The Algorithm

4.1 Motivation

Let ϕ be an automorphic (eigen) newform for the congruence subgroup Γ0(N) with rational

Hecke eigenvalues. We saw in Chapter 2 that there exists an elliptic curve Eϕ semistable at deﬁned over K = F (T ) with conductor N . As in the cases of complex and p-adic ∞ q ∞ uniformization ( 3.3 and 3.4), one would like to have a procedure that allows to calculate §§ the elliptic curve Eϕ by giving explicit equations over K.

Over the complex numbers (cf. 3.3) we saw that there exists an eﬃcient algorithm that § allows to calculate the Tate parameter and the periods with high accuracy and therefore the calculation of the equations of the corresponding elliptic curve over Q (cf. [Cre97]). In the p-adic case, although Greenberg ([Gre06]) devised an algorithm to calculated the Tate parameter from the corresponding automorphic form by means of certain multiplicative integral, it seems not known if one can compute from the Tate parameter an equation for the corresponding elliptic curve.

In the function field case, to find the Tate parameter we need to calculate the multiplicative integral (3.9). The problem of computing such integral up to an accuracy of M digits of exact precision is a priori of exponential complexity. In this work we follow the ideas of Darmon and Pollack ([DP06]) and Greenberg ([Gre06]) to develop an algorithm that allows us to calculate the Tate parameter up to a given accuracy in polynomial time. In their algorithm we have to replace the use of the logarithm by a different method, which leads us to compute several integrals of the form

(1 + at) dµ (t) (4.1) × ϕ Z O∞

45 4. The Algorithm

for a πF 2 [[t]]. We calculate this integral by iterating a certain Hecke operator as in ∈ q Greenberg’s work.

In this chapter we will deﬁne the functions to be integrated, the above mentioned Hecke operator, its properties and the algorithm that allows us to calculate the integral (3.9) up to a priori given accuracy M.

In the function ﬁeld case the coeﬃcients of the Tate curve are not rational in general, as in the complex case, however using some tools from the reduction of modular forms modulo

p and appropriate models for the elliptic curves, we manage to ﬁnd equations for Eϕ over K – this is done in the next chapter.

4.2 Elementary functions

Let I be the subset of N N given by I := (i, j) N N j i . Graphically, the × 0 { ∈ × 0| ≤ } elements of I are the points with integral coordinates in the encircled area in

line i=j

i . M

This set I has the properties:

I is closed under coordinate-wise addition; we denote this by I + I I. • ⊂ For any (i , j ) I, the set I contains the set • 0 0 ∈ ′ ′ ′ ′ ′ I := (i , j ) i i , j j + i i . (4.2) (i0,j0) { | ≥ 0 ≤ 0 − 0}

46 4.2. Elementary functions

line i=j

b j0 ′′ I I

i . i0 M

Figure 4.1: The set I′′

One associates to I the set

i j F := 1+ a π t a F 2 . (4.3) I  ij | ij ∈ q  (i,j) I  X∈ 

Elements of FI are called elementary functions in t. We may consider its elements as those of the group (1+ πF 2 [[π, t]], ) for which each term has the π-power bigger or equal than q × the t-power.

Lemma 4.2.1. The subset F of (1 + πF 2 [[π, t]], ) is a subgroup. I q × 1 Proof. We need to check that if f,g F , fg F and g− F . ∈ I ∈ I ∈ I i j k l For the ﬁrst statement, consider f =1+ (i,j) I aijπ t and g =1+ (k,l) I bklπ t with ∈ ∈ a and b F 2 . From the multiplication of series ij kl ∈ q P P

r s i j k l fg =1+  aijbkl π t + aijπ t + bklπ t

(r,s) N N0 (i,j) I, (k,l) I, (i,j) I (k,l) I X∈ ×  ∈X ∈  X∈ X∈  i+k=r, j+l=s    and since I + I I we have that (r, s) I and then fg F follows. ⊂ ∈ ∈ I

47 4. The Algorithm

For the second statement, we have that g is invertible in 1 + πFq2 [[π, t]]. We need to check that g 1 F . Let g 1 =1+ c πitj, then gg 1 = 1 implies − I − (i,j) N N0 ij − ∈ ∈ × P k l i j r s bklπ t + cijπ t + cijbklπ t =0. (4.4)

(k,l) I (i,j) N N0 (i,j) N N0, (k,l) I, X∈ X∈ × i+∈k=×Xr, j+l=∈s

Consider the set E := (i, j) N N I c =0 . { ∈ × 0 \ | ij 6 } We claim E to be empty. If it were not empty, as E N N has a lexicographic total ⊂ × 0 order, then there exists a (i , j ) E minimal. From the equation (4.4) the coeﬃcient of 0 0 ∈ πi0 tj0 is 0 and from the left hand side this is

ci0j0 + bklcmn =0. (4.5)

(m,n) N N0, (k,l) I, ∈ × ∈ m+k=Xi0, n+l=j0

If some c = 0, then since k > 1, m < i which implies that (m, n) E, and contra- mn 6 0 ∈ dicts the minimality of (i , j ). Hence (4.5) becomes c and so (i , j ) E which is a 0 0 i0j0 0 0 6∈ contradiction.

Now deﬁne a b Γ0( ) := γ = M2(O ) c 0 (mod π) ,d O× and det(γ) O 0 . ∞ c d ∈ ∞ ≡ ∈ ∞ ∈ ∞ \{ }

Note that Γ ( ) is not the Iwahori subgroup, it was deﬁned to contain at least the Iwahori 0 ∞ subgroup and matrices of the form ( π a ). However, Γ ( ) is a semigroup since it is closed 0 1 0 ∞ under the usual multiplication. Namely, let

a b a1 b1 aa1 + cc1 ab1 + bd1 γ = , γ1 = Γ0( ) and γγ1 = c d c1 d1 ∈ ∞ ca1 + dc1 cb1 + dd1

then ca1 + dc1 0 (mod π) since c 0, c1 0 (mod π). Also det(γγ1) O 0 . ≡ ≡ ≡ ∈ ∞ \{ } It remains to prove that cb1 + dd1 O× which is equivalent to val(cb1 + dd1) = 0 where ∈ ∞ val = v1/T . From the properties of the valuation one sees that

val(cb + dd ) inf val(cb ), val(dd ) 1 1 ≥ { 1 1 }

48 4.2. Elementary functions

but val(dd ) = 0 and val(cb ) 1, therefore val(cb + dd ) = 0 which implies γγ Γ ( ) 1 1 ≥ 1 1 1 ∈ 0 ∞ as desired.

The group F is equipped with a left action of Γ ( ) deﬁned by I 0 ∞

1 (κ f)(t) := f(κ− t ) for κ Γ ( ) and f F (4.6) ∗ h i ∈ 0 ∞ ∈ I where is the usual Moebious trasnformation. h·i Lemma 4.2.2. The action deﬁned in (4.6) is well deﬁned.

Proof. We need to check that for f F and κ Γ ( ) we have (κ f) F . ∈ I ∈ 0 ∞ ∗ ∈ I i j 1 a b Set f =1+ (i,j) I aijπ t and κ− = ( c d ). Then from the deﬁnition of the action we ∈ i at+b j F have (κ f)(tP)=1+ (i,j) I aijπ ct+d . So(κ f)(t) I if and only if for each term ∗j ∈ ∗ ∈ πi at+b the exponents (i, j) I. Hence it is enough to check it for πitj, (i, j) I. ct+d P ∈ ∈ First it is convenient to write

at + b 1 =(at + b) c ct + d d 1 ( − )t − d 1 =(a′t + b′) 1 πc t − ′ a b c where a′ = d , b′ = d and c′ = πd . Therefore j i at + b i j j π = π (a′t + b′) (1 πc′t)− ct + d − ∞ i j n j n = π (a′t + b′) ( c′) − (πt) − n n=0 X ∞ j n j i+n n =(a′t + b′) ( c′) − π t − n n=0 j X ∞ j k j k n i i+n k = (a′t) (b′) − ( c′) − π t k − n k=0 n=0 X X j ∞ n j i+n n j k k j k = ( c′) π t (a′) t (b′) − − n k n=0 k=0 ! X jX ∞ n j i+n k k+n j k = ( c′) π (a′) t (b′) − − n n=0 ! X Xk=0

49 4. The Algorithm

j ∞ k j k n j i+n k+n = (a′) (b′) − ( c′) π t , − n n=0 ! X Xk=0

ﬁnally if we take i′ = i + n and j′ = k + n we get the conditions given for the set I′′ and the lemma holds.

Remark 4.2.3. One can deﬁne F for arbitrary subsets I N N . It is a group whenever I ⊂ × 0 I + I I and is stable under the action of Γ ( ) if I I′′. ⊂ 0 ∞ ⊂

4.3 A Hecke operator

Fix a positive integer M. We want to compute the integral (4.1) to a precision πM . For a ﬁxed k M we deﬁne the multiplicative group ≤ k 1+ π Fq2 [[π]] Uk,M := M+1 . 1+ π F 2 [[π]] q

Note that if k′ k then U ′ U . ≤ k ,M ⊃ k,M

An important and crucial property of FI is the following, which will be proved in the Appendix 7.

Lemma 4.3.1. Given any function f F and any integer M > 1, then there exists a ∈ I ﬁnite set of indices J I and m 1, 2, ..., p 1 such that ⊆ ijδ ∈{ − } f (1 + ξδπitj)mijδ (mod πM+1) ≡ (i,j) J Y∈ where ξ F 2 is a primitive element for the extension over F , δ 0, ..., d 1 with d ∈ q p ∈ { − } the degree of the extension Fq2 over Fp. The representation is unique modulo p powers of i j fij =1+ π t , so the set

B := 1+ ξδπitj (i, j) F , i M, δ =0, ..., d 1 . M { | ∈ I ≤ − } is a multiplicative pseudo basis.

i j Lemma 4.3.1 says that the representation is unique modulo p powers of fij =1+ π t , this means that if f is a factor of a function f with p gcd(i, j), then the p-th roots of f are ij | ij in BM and they give “other” representation of f.

50 4.3. A Hecke operator

Remark 4.3.2. Let now f =1+ ξδπitj B . From the deﬁnition of the integral ij ∈ M

δ i j M+1 δ i j µ(Ul) M+1 (1 + ξ π t ) dµ (mod π ) = lim (1 + ξ π tl ) (mod π ), ×O∞ −→l tl Ul O∞ Z ∈ Y⊂ we can interchange modulo with the integral since the powers in the product do not decrease, in particular the integral is continuous with respect to the π-adic topology. We obtain that (1 + ξδπitj) dµ (mod πM+1) U . ×O∞ ∈ i,M R This leads us to deﬁne the following sets. For an integer k> 1 we consider

I := (i, j) N N i k . k { ∈ × 0| ≥ }

Note that by Remark 4.2.3 the set FI I is a group, moreover it is a subgroup of FI . ∩ k

Remark 4.3.3. As in Lemma 4.2.2 one proves that κ FI I FI I for all κ Γ0( ). ∗ ∩ k ⊂ ∩ k ∈ ∞

Let us deﬁne also the set

F : F U F is a group homomorphism I −→ 1,M Hom′(F , U ) :=   . I 1,M  s.t. F (FI I ) Uk,M , for all k 1   ∩ k ⊆ ≥ 

  It is worth pointing out that the condition on F only makes sense for k M and ≤ F (FI I ) Uk,M means that we evaluate in elements f FI such that val(f 1) k and ∩ k ⊆ ∈ − ≥ its image under F has the property val(F (f) 1) k. Since the group I acts on F by − ≥ I Moebius transformations and on Uk,M trivially, we may also deﬁne a left action of I on

Hom′(FI , U1,M ) by

1 (κ F )(f) := F (κ− f) for κ I and f F . ∗ ∗ ∈ ∈ I Lemma 4.3.4. The previous action is well deﬁned.

Proof. Let κ and κ I then we have that 1 2 ∈ 1 1 ((κ κ ) F )(f)= F ((κ− κ− ) f) 1 2 ∗ 2 1 ∗ 1 1 = F ((κ− (κ− f)) 2 ∗ 1 ∗ 1 =(κ F )(κ− f) 2 ∗ 1 ∗ = κ (κ F )(f). 1 ∗ 2 ∗

Also, using Remark 4.3.3, κ maps Hom′(FI , U1,M ) to itself. And so the lemma holds.

51 4. The Algorithm

Let Γ be a congruence subgroup of P GL2(K ). Unless otherwise stated we will take Γ to ∞ be Γ (N) for some non zero ideal N F [T ] and deﬁne the set 0 ⊂ q

′ ′ S(Γ,Hom (FI, U1,M )) := φ :Γ P GL2(K ) Hom (FI , U1,M ) φ is I-equivariant . \ ∞ −→ (4.7)

For simplicity of notation we will denote this set by S. The I-equivariance means that for

any γ P GL2(K ) and κ I, we have ∈ ∞ ∈

φ(Γγκ)(f)(t)= φ(Γγ)(κ f)(t). ∗ Writing (κ⋆φ)(Γγ) := φ(Γγκ), we can also see I-equivariance as

(κ⋆φ)(Γγ)(f)= φ(Γγ)(κ f). ∗

Remark 4.3.5. a) From Chapter 3 we have that the group GL2(K ) acts on the space ∞ 1 M0(X, Z) so the integral f d(γ− µ) makes sense for any γ GL2(K ) and ×O∞ ∗ ∈ ∞ f F . ∈ I R 1 M+1 ′ F b) The integral O∞ f d(γ− µ) (mod π ) lies in Hom ( I, U1,M ) (cf. 4.3.2) then × 1∗ M+1 the map γ d(γ− µ) (mod π ) is in S, so S is not empty. 7→R×O∞ ∗ R c) From the I-equivariance we have that any F S is uniquely determined by its ∈ values on any set of representatives (a “fundamental domain”) of the double class

Γ P GL2(K )/I, which is in canonical bijection with the edges of the quotient tree \ ∞ Γ T. \

From now on we write φ(γ) instead of φ(Γγ).

The set S is endowed with an action of the Hecke operator

π a (U φ)(γ)(f(t)) := φ γ f(πt + a) ∞ 0 1 a Fq Y∈

for f FI and γ Γ P GL2(K ). ∈ ∈ \ ∞

Lemma 4.3.6. The Hecke operator U is well deﬁned. ∞

52 4.3. A Hecke operator

π a I I ′ ′ I Proof. We define τa =( 0 1 ) with a Fq. Then for all a Fq we have τa = a Fq τa , so ∈ ∈ ∈ that for each a F and κ I we can find unique a′ F and κ′ I such that κτ = τ ′ κ′. ∈ q ∈ ∈ q ∈ ` a a Moreover for κ fixed and a variable, the map a′ a is a bijection of F . 7−→ q Let φ S, we need to check that U φ is in S, i.e., U φ is I-equivariant. Let γ ∈ ∞ ∞ ∈ Γ P GL2(K ), f FI and κ I. Then applying the U operator and using that φ is \ ∞ ∈ ∈ ∞ I-equivariant we get

(κ ⋆ U φ)(γ)(f)=(U φ)(γκ)(f) ∞ ∞ = φ(γκτ )f(τ t ) a ah i a Fq Y∈ = φ(γτ ′ κ′)f(τ t ) a ah i a Fq Y∈ 1 = φ(γτ ′ )f(τ κ′− t ) a a h i a Fq Y∈ 1 = φ(γτ ′ )f(κ− τ ′ t ) a a h i a Fq Y∈ = φ(γτa′ )(κ f)(τa′ t ) ′ ∗ h i a Fq Y∈ =(U φ)(γ)(κ f). ∞ ∗

1 Γ We recall that any Γ-invariant harmonic cocycle ϕ gives rise to a measure µϕ on M0(P (O ), Z) ∞ and conversely ϕ can be recovered from such a measure µ. From now on we will denote by µ the measure associated to ϕ H (T, Z)Γ. ϕ ∈ ! We have the following map

Γ ′ ϕ H!(T, Z) S(Γ,Hom (FI, U1,M )) ∈ ϕ −→ Φ 7−→ µϕ 1 deﬁned by Φ (γ) := d(γ− µ). µϕ ×O∞ ∗

Lemma 4.3.7. The mapR Φµϕ deﬁned above is I-equivariant.

Proof. The map Φµϕ is well deﬁned since it deﬁnes an homomorphism FI U1,M (cf. ′ → Remark 4.3.5 b)), then Φµϕ lies in Hom .

Let γ Γ P GL2(K ), f FI and κ I. We need to check the equality ∈ \ ∞ ∈ ∈ Φ (γκ)(f) = Φ (γ)(κ f). (4.8) µϕ µϕ ∗

53 4. The Algorithm

By the deﬁnition of the map and the change of variables formula we have

1 Φ (γκ)(f)= f(t) d((γκ)− µ)(t) µϕ × ∗ Z O∞ 1 1 1 = κ− (κ f)(t) d(κ− (γ− µ))(t) × −1 ∗ ∗ ∗ ∗ Z κ (κO∞) 1 = (κ f)(t) d(γ− µ)(t) × ∗ ∗ Z κO∞ 1 = (κ f)(t) d(γ− µ)(t) × ∗ ∗ Z O∞ = Φ (γ)(κ f). µϕ ∗ In the fourth equality we use the fact that κO = O , it can be seen from the identiﬁcation ∞ ∞ of the open O with the ends passing trough the edge e0 and the Iwahori subgroup stabilizes ∞ the lattice class corresponding to e0.

A crucial property of the function Φµϕ is that it is an eigenfunction of the Hecke operator.

Proposition 4.3.8. The functions Φµϕ are eigenfunctions of U with eigenvalues 1, i.e. ∞

U Φµϕ = Φµϕ . ∞

Proof. Let f be a function on FI and Φµϕ as deﬁned above. Then by deﬁnition of U we ∞ have

1 (U Φµϕ )(f)= Φµϕ (γτa)(τa− f) ∞ ∗ a Fq Y∈ 1 = f(τa t ) d((γτa)− µ)(t) ×O∞ h i ∗ a Fq Y∈ Z 1 1 1 = (τa− f)(t) d(τa− γ− µ)(t) −1 ×τa (τaO∞) ∗ ∗ ∗ a Fq Y∈ Z 1 = f(t) d(γ− µ)(t) ×τa(O∞) ∗ a Fq Y∈ Z 1 = f(t) d(γ− µ)(t) ×a+πO∞ ∗ a Fq Y∈ Z 1 = f(t) d(γ− µ)(t) ×˙ ∈F (a+πO∞) ∗ Z ∪a q 1 = f(t) d(γ− µ)(t) × ∗ Z O∞

= Φµϕ (f).

54 4.3. A Hecke operator

The important thing to note here is that the set 1 + πFq2 [[π]] can be seeing as the subset

FI0 of FI for I0 := N 0 . The set FI0 is preserved under the action of I since the action ×{ } ′ is trivial. Given any f Hom (F , U ) we may restrict it to F and therefore induce a ∈ I 1,M I0 restriction map

′ ′ Hom (F , U ) res Hom (F , U ) I 1,M −→ I0 1,M f f F . 7−→ | I0

This map induces naturally another restriction map res*, which by abuse of notation will be also denoted by res, between the following spaces

′ ′ S(Γ,Hom (F , U )) res S(Γ,Hom (F , U )). I 1,M −→ I0 1,M

Finally consider the map Φ0,µϕ deﬁned as follows T Γ S ′ F ϕ H!( , Z) (Γ,Hom ( I0 , U1,M )) ∈ ϕ −→ Φ , 7−→ 0,µϕ

−1 (γ µϕ)(O∞) M+1 where Φ (γ)(f) := f ∗ mod π , for f F . 0,µϕ ∈ I0

The map is well deﬁned since f FI0 is constant, and then its integral is actually −1 ∈ γ µϕ(O∞) ϕ(γe0) f ∗ = f , where e0 is the standard edge.

These considerations lead us to Lemma 4.3.9. The following diagram is commutative.

Γ H!(T, Z) ❲❲❲ ❲❲❲❲ Φ ❲❲❲❲❲ 0,µ∗ Φµ∗ ❲❲❲❲❲ ❲❲❲❲❲+ ′ res / ′ S(Γ,Hom (FI, U1,M )) S(Γ,Hom (FI0 , U1,M )).

Proof. The proof is straightforward. We need to check that res Φ = Φ holds for all ◦ µϕ 0,µϕ ϕ H (T, Z)Γ. ∈ ! 1 In order to calculate res Φ , we need to evaluate f d(γ− µ)(t) for f F and ◦ µϕ ×O∞ ∗ ∈ I0 since f is constant R − 1 (γ 1 µ)(O∞) f d(γ− µ)(t)= f ∗ . × ∗ Z O∞ M+1 Taking modulo π this is precisely Φ0,µϕ (f).

55 4. The Algorithm

Now, we are ready to describe the algorithm that allows us to calculate the integral f dµ(t) for f F . ×O∞ ∈ I ′ SupposeR that we have φ S(Γ,Hom (F , U )) such that it is an eigenfunction for the ∈ I 1,M Hecke operator U with eigenvalue 1 and also that res φ = φ0 (one candidate for φ is Φµϕ ∞ ◦ for some Γ-invariant harmonic cocycle ϕ) and we want to evaluate φ at 1+ ξδπM tj B . ∈ M Then for any γ Γ we have ∈ δ M j δ M j (U φ)(γ)(1 + ξ π t )= φ(γτa)(1 + ξ π (a + πt) ) ∞ a Fq Y∈ M δ j = φ0(γτa)(1 + π ξ a ). (4.9)

a Fq Y∈

In particular, when φ is Φµϕ the last product equals

(1 + πM ξδaj)ϕ(γτae0)).

a Fq Y∈ We compute now

δ M 1 j δ M 1 j U (φ)(γ)(1 + ξ π − t )= φ(γτa) 1+ ξ π − (a + πt) ∞ a Fq Y∈ j δ M 1 j n j n = φ(γτ ) 1+ ξ π − (πt) a − a n a Fq n=0 ! Y∈ X δ M 1 j j 1 j j = φ(γτa) 1+ ξ π − (a + jπta − + ... + π t ) .

a Fq Y∈ δ M 1 j δ j 1 M M+1 = φ(γτa) (1 + ξ π − a )(1 + ξ ja − π t)(1 + O(π )) .

a Fq Y∈

δ M 1 j From the previous calculation we see that the value of φ(γτa)(1+ ξ π − a ) is easy to cal- δ M j 1 culate since it does not depend on t. The calculation of the value of φ(γτa)(1+ξ π a − jt) reduces hence to the previous case (4.9). The above method can be continued inductively.

Summarizing, given any function f F , M > 1 and ϕ H (T, Z)Γ, to calculate the ∈ I ∈ ! integral f dµ up to accuracy of M digits of exact precision, we decompose f as ×O∞ ϕ

R mij f f (mod πM+1) ≡ ij f B ijY∈ M δ i j where fij =1+ ξ π t .

56 4.3. A Hecke operator

So it is enough to know the value of the integral at the functions fij, that is the value of

Φµϕ (γ)(fij) for all fij BM and all γ Γ P GL2(K ). By Remark 4.3.5 we only need to ∈ ∈ \ ∞ calculate this value for each f B and ﬁnitely many. Hence as a ﬁrst step, we produce ij ∈ M

tables with all the values for Φµϕ (γ)(fij). In the following lines we describe the algorithm which computes such table.

δ i j δ′ i′ j′ Let us deﬁne an order relation on BM . We say that 1 + ξ π t > 1+ ξ π t if and only if their exponents satisfy one of the following two conditions:

1) If j > 0 and j′ > 0

1.1) i < i′ or

1.2) (i = i′ and j < j′) or

1.3) (i = i′ and j = j′ and δ′ 6 δ).

2) If j′ =0

2.1) j > 0 or

2.2) (j = 0 and i < i′) or

2.3) (j = 0 and i = i′ and δ 6 δ′).

Note that we can induce a partition of the set BM in the following way. For a ﬁxed δ we deﬁne the following subset of BM

B := f B f =1+ ξδπitj (4.10) M,δ { ∈ M | }

Clearly BM = ˙ δ 0,...,d 1 BM,δ. Note also that we can identify the elements of BM,δ with ∪ ∈{ − }

57 4. The Algorithm

the points with integer coordinates of the graph

line i=j

BM,δ

i M

We will see later how to use this representation to explain the algorithm.

From the deﬁnition of the operator U , given a representative γ of a class in Γ P GL2(K )/I, ∞ \ ∞ if e is the edge in T corresponding to [γ]1, we have that γτa represents all the edges in T with terminal o(e).

Remark 4.3.10. For γ and e as above we have that the classes of γτ for a F in the a ∈ q quotient graph are identiﬁed, so we do not need to calculate the integral q times.

If we apply the operator M times, then we move in the tree T distance M from the starting edge. So we need representatives for the quotient tree up to level M. Let us denote by R such a set of representatives.

For each γ R we need to calculate the integral Φ (γ)(f ) for all f B . To each γ ∈ µϕ ij ∈ M we attach d “triangular matrices” Tγ,δ whose entries are the values of the integral at all δ i j functions f of BM,δ, that is Tγ,δ[i, j] = Φµϕ (γ)(1 + ξ π t ).

Observe that the constants corresponds to the points located over the i-axis of the graphic representation of BM,δ . We start by ﬁlling the tables Tγ,δ with the value Φµϕ (γ)(f) for f B constant for all δ 0, 1, ..., d 1 . ∈ M,δ ∈{ − } First for γ R we calculate the value ϕ(γe ) and then Φ (γ)(f)= f ϕ(γe0). We store this ∈ 0 µϕ value at the corresponding place of Tγ,δ. That is at ﬁrst stage all the values for functions

58 4.3. A Hecke operator

over the axis i of B are known for all γ R and δ 0, 1, ..., d 1 , this corresponds to M,δ ∈ ∈{ − } the last ﬁle of the tables Tγ,δ, since we can identify the entries of the table with the points

of BM,δ.

Remark 4.3.11. When we apply the U to Φµϕ evaluated at a representative γ we move ∞ with τa to other representatives, so we need to ﬁll the tables Tγ,δ simultaneously for a ﬁxed f B and running over γ R. ∈ M ∈

We start with the smallest f B and calculate Φ (γ)(f), that is we start by ﬁlling the ∈ M µϕ table Tγ,δ by the upper right corner. Since the power of π and t is M, when we apply the operator, it reduces immediately to constants. Whose values were already calculated.

Let suppose that f = 1+ ξδπi0 tj0 B and we know the value of Φ (γ)(g ) in all i0j0 ∈ M µϕ ij g B with g > f . We can interpret this situation with a graphic in the following ij ∈ M ij i0j0 way j

line i=j

b j0 (i0, j0)

BM,δ

i i0 M

were the shaded area1 represents the functions f B for which we know the value of ∈ M,δ Φµϕ (γ)(f) and the dashed blue line represents the functions in BM,δ with exponent i0 in π to be integrated.

Applying the U operator to Φµϕ (γ)(fi0j0 ) we have ∞

U (Φµϕ (γ)(fi0j0 )) = Φµϕ (γτa)(fi0j0 (πt + a)). ∞ a Fq Y∈ 1Although is a discrete set we use solid color to understand better the method.

59 4. The Algorithm

Writing for every a F γτ = σ γ κ with σ Γ, γ R and κ I and using the ∈ q a a a a a ∈ a ∈ a ∈ Iwahori equivariance of Φµϕ and the Γ-invariance we have e e

Φµϕ (γτa)(fi0j0 (πt + a))=Φµϕ (σaγaκa)(fi0j0 (πt + a)) 1 = Φµϕ (γa)(fi0j0 )(τaκa− t ) e h i δ i0 1 j0 = Φµϕ (γa)(1 + ξ π (τaκa− t ) . e h i e δ i0 1 j0 Now we need to decompose the function 1 + ξ π (τ κ− t ) as a product of elements in a a h i the pseudo-basis BM

δ i0 1 j0 mij M+1 1+ ξ π (τ κ− t ) f (mod π ). a a h i ≡ ij f B ijY∈ M

This decomposition has the property that each fij in it satisﬁes fij > fi0j0 , moreover the ′′ exponents of fij are in the set I (see ﬁgure 4.1), then we know its integral which is stored in one of the tables corresponding to γa.

Remark 4.3.12. At (i0, j0) need onlye values in the blue shaded area.

Example 4.3.13. Let N = T 3 over F , in Chapter 2 2.8 we showed the corresponding 2 § graph. Let ϕ be the unique harmonic cocycle with rational Hecke eigenvalues (observe that 0 1 there is only one cycle in the quotient graph) and let γ = 2 R be the representing 1 T +T ∈ matrix of the edge e = (2, 5). Working with M = 7 and f =1+ π3t2 B , to calculate ∈ 7 Φµϕ (γ)(f) we apply the operator U . ∞

3 2 3 U (Φµϕ (γ)(1 + π t )) = Φµϕ (γτa)(1 + π (πt + a)). ∞ a F2 Y∈ We need to decompose the matrices γτ0 and γτ1 as γaγaκa, so we have

0 1 γτ0 = 1/T T 2+T e T 2+T +1 1 T 2 T 3+T 1/T 0 = T 3 T+1 T 4+T 3+T 2+1 T 5+T 4+T 2 1/T 2 1/T T 2 T 3+T and the matrix γ0 = T 4+T 3+T 2+1 T 5+T 4+T 2 is a representative for the edge (7, 2) and e 1/T 0 1 0 1/T 0 κ0 = 1/T 2 1/T = 1/T 1 0 1/T .

60 4.4. The change of variables and calculation of the integral

1 0 We can take κ0 to be 1/T 1 then

3 1 2 5 2 7 4 9 6 1+ π (τ κ− t ) =1+ π t + π t + π t + ... 0 0 h i (1 + π5t2)(1 + π7t4) (mod π8). ≡ 5 2 7 4 3 2 Both functions (1 + π t ) and (1+ π t ) belong to B7,0 and are smaller than f =1+ π t ,

so its integral is stored in the table Tγ0,0, moreover they are Tγ0,0[5, 2] and Tγ0,0[7, 4], e e e respectively.

Analogously the matrix γτ1 may be decomposed as above, in this case we have that the 1 0 representative corresponds to the edge (4, 2) and κ1 = 1/T 1 , we have

3 1 2 3 5 2 7 4 9 6 1+ π (τ κ− t ) =1+ π + π t + π t + π t ... 1 1 h i (1 + π3)(1 + π5t2)(1 + π7t4) (mod π8). ≡ 3 5 2 7 4 Again the values of (1+π ),(1+π t ) and (1+π t ) are Tγ1,0[3, 0], Tγ1,0[5, 2] and Tγ1,0[7, 4], e e e respectively.

4.4 The change of variables and calculation of the integral

In this section we will see that the calculation of the multiplier cα(γ) deﬁned in Chapter 3,

reduces to integrate functions on FI . Recall that cα(γ) is deﬁned by the following integral

t γz − 0 dµ (t) for z Ω. (4.11) × t z ϕ 0 ∈ Z∂Ω − 0 Using the partition induced from the ends, we will break up the domain of integration in a ﬁnite union of disjoint open compacts and then by the change of variables formula, transform each integral resulting from the partition, in one of the form

f d(µ)(t) × Z O∞ for some f F . ∈ I Before describing the change of variables (since the integral (4.11) does not depend on

the choice of z0) we will describe brieﬂy in Subsection 4.4.1 a way to construct a suitable

z0 using the reduction map. After this, in Subsection 4.4.2, we construct explicitly the partition of ∂Ω induced from a ﬁxed edge e of the tree.

61 4. The Algorithm

4.4.1 Choosing the z0

Let ξ be a generator of a normal basis of Fq2 over Fp, then ξ is in the standard aﬃnoid z C z 1 z c 1 c F , i.e. it is a lifting of the standard vertex under the { ∈ || | ≤ | − | ≥ ∀ ∈ q} reduction map, that is λ(ξ) = v0. So we may take z0 = ξ. Moreover, given any vertex v

on the tree T, we know that there is a γ GL2(K ) such that γv0 = v, where v0 is the ∈ ∞ standard edge. From the GL2(K )-invariance of the reduction map, we have that z = γz0 ∞ is a lifting of the vertex v under the reduction map .

πk u It is straightforward to prove that given any vertex v and γ = 0 1 the normal form representing v, lifting to Ω is just giving by applying γ to ξ as a Moebius transformation, namely, z = ξπk + u, and so val(z)

Let z and w be two diﬀerent points in Ω such that λ(z) = λ(w) then we say that they are 6 consecutive if λ(z) and λ(w) are consecutive vertices in the tree.

4.4.2 The partition of the border

Let e be any edge of the tree, we know from (2.4) that it induces a partition of ∂Ω. In order to construct explicitly such partition, it is convenient to write the vertices that determine e in normal form. So without loss of generality we may assume that the vertices v and k v′ are represented by v = [k,u] and v′ = [k +1,u + aπ ] for a F , respectively. Also ∈ q suppose that v is closer than v to the “line” A(0, ) in other words, the edge (v′, v) 1 ∞ points to inﬁnity. We know from Lemma 2.4.5 that all the neighbors of v diﬀerent from k v′ are given by [k +1,u + aπ ] for a F a and the neighbors of v′ are of the form ∈ q \ 0 [k +2,u + a πk + bπk+1] with b F . Graphically we have the following 0 ∈ q k k k+1 [k +1,u + aπ ] [k +2,u + a0π + bπ ] b b

b b q 1 k − [k,u] [k +1,u + a0π ] edges edges . bb b . q . .

b b

b k 1 b [k,u mod π − O ] ∞

62 4.4. The change of variables and calculation of the integral

From the Lemma 2.5.4 we have that to a vertex v of the form [k,u] we attach the open set u + πkO which corresponds to the ends associated to edges e having v as terminal. ∞ Therefore u + πkO = (u + aπk + πk+1O ) ∞ ∞ a Fq [∈ and one can easily observe

k+1 Lemma 4.4.1. Given two neighbor vertices [k,u] and [k +1,u + a0π ] with k u O (mod π ) and k > 0 and a0 Fq, then ∈ ∞ ∈ k k+1 k k+1 k+2 1 k u + aπ + π O ˙ u + a0π + bπ + π O ˙ P (K ) u + π O a Fq,a=a0 b Fq ∞ ∈ 6 ∪ ∞ ∈ ∪ ∞ \ ∞ is the partition of ∂Ω induced from the identiﬁcation given by then ends.

Finally, in this partition we can identify three kinds of open sets:

k k+1 i) q 1 of the form u + aπ + π O a Fq, a = a0, which we denote by Wa. − ∞ ∈ 6 k k+1 k+2 ii) q of the form u + a0π + bπ + π O , b Fq, which we denote by Wa0,b. ∞ ∈ iii) One of the form P1(K ) (u + πkO ) denoted by W . ∞ \ ∞ ∞

4.4.3 The change of variables

Let ϕ H (T,G)Γ0(N), let µ be the corresponding measure associated to it and let α Γ ∈ ! ϕ ∈ be a lifting of ϕ, that is, j(α)= ϕ. Consider also v0, v1, ..., vr to be a path in the tree which is the lifting of a cycle c of the quotient tree Γ T. Then there exists a γ Γ such that \ ∈ γv0 = vr and let z0, z1, ..., zr = γ(z0) be consecutive points on Ω above v0, v1, ..., vr. Since the integral is multiplicative, we have

z t c (γ)= 0 − dµ (t) α × γz t ϕ Z∂Ω 0 r 1 − − z t = i − dµ (t) × z t ϕ Z∂Ω i=0 i+1 − r 1 Y − z t = i − dµ (t). × z t ϕ i=0 ∂Ω i+1 Y Z −

63 4. The Algorithm

Therefore, we will concentrate in how to calculate an integral of the form

z t i − dµ(t). × z t Z∂Ω i+1 −

Using the partition of the border from Lemma 4.4.1, we can break up the integral in 2q integrals:

f(t) dµ(t)= f(t) dµ(t) f(t) dµ(t) f(t) dµ(t) ×∂Ω ×Wa ×W ×W∞ a Fq,a=a0 b Fq a0,b Z ∈ Y6 Z Y∈ Z Z where f(t)=(z t)/(z t). i − i+1 −

Integrating over Wa

Consider the map

k k+1 γ : O Wa = u + aπ + π O ∞ −→ ∞ t γ(t)= u + aπk + tπk+1. 7−→

πk u+aπk+1 It is given by the invertible Moebius transformation 0 1 , and hence a bijection. By abuse of notation we also write γ for this matrix. Note that the matrix γ represents one of the neighbors of the vertex v.

By the previous lines, the inverse of the map γ is given by the inverse of the matrix γ acting on Wa by Moebius transformations i.e.,

1 γ− : Wa O −→ ∞ k (k+1) t a + uπ− + π− t. 7−→

64 4.4. The change of variables and calculation of the integral

We make now the change of variables,

zi t zi γ(t) 1 − dµ(t)= − d(γ− µ)(t) × z t × z γ(t) ∗ Z Wa i+1 − Z O∞ i+1 −

k k+1 zi (u + aπ + tπ ) 1 = − d(γ− µ)(t) × z (u + aπk + tπk+1) ∗ Z O∞ i+1 −

k k+1 zi u aπ tπ 1 = − − − d(γ− µ)(t) × z u aπk tπk+1 ∗ Z O∞ i+1 − − −

k k+1 k (zi u aπ )(1 tπ /(zi u aπ )) 1 = − − − − − d(γ− µ)(t) × (z u aπk)(1 tπk+1/(z u aπk)) ∗ Z O∞ i+1 − − − i+1 − −

k+1 k µ(γO∞) tπ 1 zi u aπ O∞ 1 z u aπk d(γ− µ)(t) = × − i− − ∗ . − − k tπk+1 zi+1 u aπ 1 RO∞ 1 z u aπk d(γ− µ)(t) − − × − i+1− − ∗ R πk+1 The quantity k has positive valuation since val(z ) 6 k hence the function (zi+1 u aπ ) i πk+1 − − F πk+1 F 1 (z u aπk) t is an element of I . Analogously, 1 (z u aπk) t I . − i− − − i+1− − ∈

Integrating over W ∞

We will write the corresponding change of variables as a composition of two maps. In order to construct the ﬁrst one, note that the following open sets are canonically isomorphic

πkO u + πkO ∞ −→ ∞ t u + t, 7−→

u + πkO πkO ∞ −→ ∞ t′ u t′ − 7−→

where u is thesame as in the case Wa. These two maps are preserved under complement with respect to P1(K ), ∞ P1(K ) πkO P1(K ) (u + πkO ) ∞ \ ∞ −→ ∞ \ ∞ t u + t, 7−→

65 4. The Algorithm

P1(K ) (u + πkO ) P1(K ) πkO ∞ \ ∞ −→ ∞ \ ∞ t′ u t′. − 7−→

For the second one, let us now consider a map from P1(K ) πkO as follows ∞ \ ∞

P1(K ) πkO O ∞ \ ∞ −→ ∞ k 1 t π − /t. 7−→

This map is a bijection and is well deﬁned since if t P1(K ) πkO then val(t) 6 k 1 ∈ ∞ \ ∞ − k 1 > 0 πk−1 and val(π − /t) 0. As a Moebius transformation the map is given by the matrix 1 0 0 1 whose inverse is πk−1 0 . So the inverse of the map above is O P1(K ) πkO ∞ −→ ∞ \ ∞ (k 1) t 1/tπ− − . 7−→

Then we get the change of variables that we are interested in by composing the maps in the diagram

W P1(K ) πkO O ∞ −→ ∞ \ ∞ −→ ∞ πk 1 t t u − . 7−→ − 7−→ t u − In summary we have that the change of variables is given by the following map

1 γ− : W O ∞ −→ ∞ πk 1 t − 7−→ t u − and its inverse given by

γ : O W ∞ −→ ∞ k 1 π − t′ + u. 7−→ t′

66 4.4. The change of variables and calculation of the integral

u πk−1 Finally applying the formula of change of variables with γ = 1 0 , we have

zi t zi γ(t) 1 − dµ(t)= − d(γ− µ)(t) × z t × z γ(t) ∗ Z W∞ i+1 − Z O∞ i+1 −

πk−1 zi ( t + u) 1 = − d(γ− µ)(t) πk−1 ×O∞ z ( + u) ∗ Z i+1 − t

k 1 tzi π − tu 1 = − − d(γ− µ)(t) × tz πk 1 tu ∗ Z O∞ i+1 − − −

k 1 (zi u) − π − (1 t πk−1 ) 1 = − d(γ− µ)(t) k 1 (zi+1 u) ×O∞ π (1 t −− ) ∗ Z − − πk 1

zi u 1 1 k−−1 t d(γ− µ)(t) = ×O∞ − π ∗ . zi+1 u 1 1 −− t d(γ µ)(t) ×RO∞ − πk 1 − ∗ R

zi u zi+1 u Again −− and −− have both positive valuation since val(z u) > k and πk 1 πk 1 i − zi u zi+1 u val(z u) >k. Hence the functions 1 −− t and 1 −− t are in F . i+1 − − πk 1 − πk 1 I

Integrating over Wa0,b

Consider the map

γ : O Wa0,b ∞ −→ t γ(t)= u + a πk + bπk+1 + tπk+2. 7−→ 0

This map is a change of variables, the proof is similar as the one for the open Wa. So making the change of variable we have

67 4. The Algorithm

zi t zi γ(t) 1 − dµ(t)= − d(γ− µ)(t) ×W zi+1 t ×O∞ zi+1 γ(t) ∗ Z a0,b − Z −

k k+1 k+2 zi (u + a0π + bπ + tπ ) 1 = − d(γ− µ)(t) × z (u + a πk + bπk+1 + tπk+2) ∗ Z O∞ i+1 − 0

k k+1 k+2 zi u a0π bπ tπ 1 = − − − − d(γ− µ)(t) × z u a πk bπk+1 tπk+2 ∗ Z O∞ i+1 − − 0 − −

k k+1 tπk+2 zi u a0π bπ 1 k k+1 (zi u a0π bπ 1 = − − − − − − − d(γ− µ)(t) k+2 ×O∞ k k+1 tπ ∗ (z u a π bπ ) 1 k k+1 Z i+1 0 z u a0π bπ − − − − i+1− − −

k+2 k k+1 µ(γO∞) tπ 1 1 k k+1 d(γ− µ)(t) zi u a0π bπ O∞ z u a0π bπ = − − − × − i− − − ∗ . k k+1 tπk+2 1 zi+1 u a0π bπ 1 k k+1 d(γ µ)(t) RO∞ z u a0π bπ − − − − × − i+1− − − ∗ R πk+2 The quantity k k+1 has positive valuation since val(zi) 6 k hence the function zi u a0π bπ πk+2 − − − πk+2 1 k k+1 t is an element of F . Analogously for 1 k k+1 t . z u a0π bπ I z u a0π bπ − i− − − − i+1− − −

Remark 4.4.2. Note that the integral over the open sets Wa and Wa0,b have positive valuation.

68 5. Applications and examples

In Chapter 4 we described an algorithm to compute the integrals needed to ﬁnd the Tate period. Here we describe how to obtain the Tate period explicitly from the integral and we

use it to obtain elliptic curves deﬁned over Fq(T ) with the desired conductor. The chapter is organized as follows, in 1-5 we make a short review of the theory of elliptic curves, §§ supersingular curves, Eisenstein series, reduction modulo p of modular forms and the Tate curve. In 6 we describe how to calculate the Tate parameter and in 7 we give algorithms § § to obtain from the Tate parameter, the equations for elliptic curves over Fq(T ).

5.1 Elliptic curves

We recall some basic facts from the theory of elliptic curves. Proofs for most of the theorems can be found in [Sil09].

2 Let K be a ﬁeld, and let us consider projective curves in PK deﬁned by

2 2 3 2 2 3 ZY + a1ZXY + a3Z Y = X + a2ZX + a4Z X + a6Z (5.1)

with coeﬃcients a K. We may usually consider the corresponding aﬃne curve for i ∈ (Z =0, x := X/Z, y := Y/Z) 6

2 3 2 y + a1xy + a3y = x + a2x + a4x + a6. (5.2)

The only missing point (0 : 1 : 0) is always smooth.

Deﬁnition 5.1.1. An elliptic curve over a ﬁeld K is a smooth projective curve E given by the equation (5.1).

69 5. Applications and examples

The equation (5.2) is called a (long) Weierstrass equation for E.

Let us deﬁne the quantities

2 b2 = a1 +4a4,

b4 =2a4 + a1a3, 2 b6 = a3 +4a6, b = a2a +4a a a a a + a a2 a2, 8 1 6 2 6 − 1 3 4 2 3 − 4 c = b2 24b , 4 2 − 4 c = b3 + 36b b 216b , 6 − 2 2 4 − 4 ∆= b2b 8b3 27b2 +9b b b , − 2 8 − 4 − 6 2 4 6 3 j = c4/∆.

Deﬁnition 5.1.2. Given a projective curve E over a ﬁeld K as above, the quantity ∆ is called the discriminant and in case ∆ = 0 the quantity j is called the j-invariant of E. 6

Observe that a projective curve deﬁned by the equation (5.1) is not singular if and only if its discriminant ∆ = 0. That is, a curve given by the equation (5.1) is an elliptic curve if 6 and only if ∆ = 0. 6 If the characteristic of K is not 2, then the change of variables y y a1 x a3 transforms 7→ − 2 − 2 a Weierstrass equation in one of the form

b b b y2 = x3 + 2 x2 + 4 x + 6 . 4 4 4

If in addition the characteristic of K is not 3 then a further change of variables x x + b2 7→ 12 gives the equation c c y2 = x3 4 x 6 . − 48 − 864 Finally with the change of variables y y/2 the equation becomes 7→ y2 =4x3 g x g (5.3) − 2 − 3 with g2 = 108c4 and g3 = 216c6.

Remark 5.1.3. The last equation is not a Weierstrass equation since the coeﬃcient of x3 is not 1. However, it is a convenient expression ( e.g. when one studies elliptic curves over C). This form will be used later in this chapter.

70 5.1. Elliptic curves

In summary, if the characteristic of K is not 2 or 3, we may and will assume that our elliptic curve has Weierstrass equation of the form

E : y2 = x3 + Ax + B (5.4)

by considering the change of variables y 2y with A = g2 and B = g3 . This form is 7→ − 4 − 4 called the (short) Weierstrass form. In the following we may simply say Weierstrass form, for short or long equation.

In characteristic 2 the discriminant and the j-invariant of an elliptic curve in Weierstrass form are given explicitly by

6 5 4 2 4 2 4 3 3 ∆= a1a6 + a1a3a4 + a1a2a3 + a1a4 + a3 + a2 + a3

and 12 j = a1 /∆.

Note that the discriminant and the j-invariant of an elliptic curve given by a short Weier- strass equation are given by the formulas

∆= 16(4A3 + 27B2) − and

64A3 j(E)= 1728 . − ∆

Two elliptic curves E and E′ deﬁned by the Weierstrass equations with variables x and y

and with variables x′ and y′, respectively, are isomorphic over K if and only if there exists r,s,t,u K with u = 0 such that the change of variables ∈ 6

2 x = u x′ + r, 3 2 y = u y′ + su x′ + t. (5.5)

The transformation in (5.5) is referred to as an admissible change of variables. Clearly, this transformation is invertible and its inverse also deﬁnes an admissible change of variables

that transforms E′ into E.

71 5. Applications and examples

Theorem 5.1.4. Let E and E′ be two elliptic curves deﬁned over an algebraically closed

ﬁeld K. Then E is K-isomorpic to E′ if and only if they have the same j-invariant.

If we apply to a Weierstrass equation the change of variables given by (5.5) the coeﬃcients of the new curve and its associated quantities (denoted with primes) are compiled in the following list:

2 ua1′ = a1 + s , 2 2 u a′ = a sa +3r s , 2 2 − 1 − 3 u a3′ = a3 + ra1 +2t, 4 2 u a′ = a sa +2ra (t + rs)a +3r 2st, 4 4 − 3 2 − 1 − 6 2 3 u a′ = a + ra + r a + r ta rta , 6 6 4 2 − 3 − 1 2 u b2′ = b2 + 12r, 4 2 u b4′ = b4 + rb2 +6r , 6 2 3 4 u b6′ = b6 +3rb6 +3r b4 + r b2 +3r , 4 u c4′ = c4, 6 u c6′ = c6, 12 u ∆′ = ∆,

j′ = j.

The group law

The set of points E(K) on an elliptic curve has a natural structure of an abelian group. The group law can be characterized in a number of equivalent ways. We characterize it by the following two rules:

1) The point O = (0 : 1 : 0) is the identity of the group.

2) If a line L intersects E in three K-points P, Q, R E(K) (taking multiplicities into ∈ account), then P + Q + R = O in the group law.

From these one can deduce

72 5.1. Elliptic curves

a) The point P is the third intersection point of the line through O and P . − b) Given P, Q E(K) not equal to O, the line through P and Q (if P = Q then take ∈ the tangent line at P ) intersects E at P, Q and a third point R E(K). If R = O, ∈ then P + Q = O; otherwise P + Q = R where R can be constructed as in a). − −

It is easy to see that, at least generically, the coordinates of P + Q can be expressed as rational functions in the coordinates of P and Q. For example, if P =(x, y) and Q =(x′,y′) are in the curve given by a Weierstrass form y2 = x3 + Ax + B, then

2 y′ y x(P + Q)= − x x′ x x − − ′ − and x4 2Ax2 8Bx + A2 x(2P )= − − . 4x3 +4Ax +4B

Singular curves

Let E be a cubic curve given by the equation (5.2) with discriminant ∆ = 0, then E has a singular point (cf. [Sil09, Prop. 1.4 a)]). Actually one can easily show, that there is only

one singular point, let say P . Let c4 be the quantity associated to the Weierstrass equation of E, there are two possibilities for the singularity at P :

1) If c = 0 then P has two distinct tangent directions. In this case P is called a node. 4 6

2) If c4 = 0 then P has only a single tangent direction. In this case P is called a cusp.

Deﬁnition 5.1.5. Let E be a (possibly singular) cubic curve given by a Weierstrass equa-

tion (5.2). The non-singular part of E, denoted by Ens, is the curve with its singular point

removed. Similarly, if E is deﬁned over K, then Ens(K) is the set of non-singular points of E(K).

The set Ens(K) has a particularly simple structure described in the following proposition.

Proposition 5.1.6. Let E be a cubic curve given by a Weierstrass equation with discriminant ∆=0 with E singular point P . Then the group law makes Ens(K) into an abelian group.

73 5. Applications and examples

a) Suppose E has a node, and let

y = α1x + β1 and y = α2x + β2

be the two distinct tangent lines to E at P . Then the map

E (K¯ ) K¯ × ns −→ y α x β (x, y) − 1 − 1 7−→ y α x β − 2 − 2 is an isomorphism (of abelian groups).

b) Suppose E has a cusp, and let y = αx + β

be the tangent line to E at P . Then the map

E (K¯ ) K¯ + ns −→ x x(P ) (x, y) − 7−→ y αx β − − is an isomorphism.

Isogenies

We turn now to the deﬁnition of morphisms between elliptic curves.

Definition 5.1.7. Let E1 and E2 be two elliptic curves over K. Let L/K be a field extension, an isogeny (over L) between E and E is a non-constant morphism φ : E 1 2 1 −→ E2 defined over L that satisfies φ(O)= O. We say that two curves E1 and E2 are isogenous if there exists an isogeny from E1 to E2.

Remark 5.1.8. If the extension L of K is not speciﬁed then we are assuming that the isogeny is deﬁned over the algebraic closure K¯ of K.

From the deﬁnition one can see that the relation of isogeny is an equivalence relation on elliptic curves. We have also

Proposition 5.1.9. Every isogeny φ : E E is a group homomorphism. 1 −→ 2

74 5.1. Elliptic curves

The set of isogenies from E1 to E2 together with the zero map is denoted by Hom(E1, E2). It is a group under addition. The group Hom(E, E) is a ring which we denote by End(E). It is called the endomorphism ring of E. The operations in End(E) are given by

(φ + ψ)(P )= φ(P )+ ψ(P ) and (φψ)(P )= φ(ψ(P )).

The invertible elements of End(E) form the automorphism group of E which is denoted by Aut(E).

Theorem 5.1.10. a) Let E1 and E2 two elliptic curves over K, then the group of iso-

genies, deﬁned over K¯ , Hom(E1, E2) is a torsion free Z-module.

b) Let E be an elliptic curve over K, then the endomorphism ring End(E) is a (not necessarily commutative) ring of characteristic 0 with non zero divisors.

c) Let E be an elliptic curve deﬁned over a ﬁeld K. Then the endomorphism ring of E is one of the following

Z, End(E)= an order in a imaginary quadratic ﬁeld,   a maximal order in a quaternion algebra.   The last case only happens if char (K)= p> 0.

Elliptic curves over ﬁnite ﬁelds

n Let E be an elliptic curve defined over a finite field Fq with q = p , p a prime. The first important result dealing with elliptic curves over finite fields is the following fact established by Lang and Weil [LW54].

Theorem 5.1.11. Any smooth cubic curve E defined over a finite field Fq has a Fq-rational point.

The set of Fq-rational points of an elliptic curve defined over a finite field Fq is finite. Hasse’s theorem on elliptic curves, also referred to as the Hasse bound, provides an estimate of the number of points on an elliptic curve over a finite field.

75 5. Applications and examples

Theorem 5.1.12 (Hasse). Let E/Fq be an elliptic curve. Then

#E(F ) (q + 1) 6 2√q. | q − |

Definition 5.1.13. Let E be an elliptic curve defied over a finite field Fq. The Frobenious endomorphism is given by

Φ : E(F ) E(F ) E q −→ q (x, y) (xq,yq). 7−→

The Frobenious endomorphism is strongly related with #E(Fq) by the following

Theorem 5.1.14. Let E be an elliptic defined over a finite field Fq and let #E(Fq) = q +1 a. Then the Frobenious endomorphism satisfies the equality − Φ2 aΦ + q =0. E − E Definition 5.1.15. The quantity a from the theorem is called the Frobenious trace.

Elliptic curves over local ﬁelds

Let K be a complete local ﬁeld with normalized valuation ν : K× Z. Let R be the −→ ring of integers of K with maximal ideal p and residue ﬁeld k = R/p. Let also ̟ be a uniformizer for R, that is p = ̟R.

2 3 2 For a given elliptic curve over K with equation y + a1xy + a3y = x + a2x + a4x + a6, 2 3 the substitution (x, y) (u− x, u− y) leads to a new equation in which each coefficient a 7→ i i is replaced by u ai. If we choose u to be divisible by a sufficiently large power of ̟, we obtain a Weierstrass equation with coefficients in R.

Definition 5.1.16. Let E/K be an elliptic curve defined by the affine Weierstrass equation

2 3 2 E : y + a1xy + a3y = x + a2x + a4x + a6 (5.6)

with a R. We say that (5.6) is a minimal Weierstrass equation for E if ν(∆) is minimal i ∈ among all Weierstrass equations deﬁning E with coeﬃcients in R.

The minimal Weierstrass equation always exists, since ν is discrete and we can choose among all Weierstrass equations with coeﬃcients in R, one that minimalizes ν(∆). If the

76 5.1. Elliptic curves

equation for an elliptic curve is not minimal, there is a coordinate change giving a new 12 equation with discriminant ∆′ = u− ∆ R. Thus ν(∆) can only be changed by adding ∈ 4 6 or subtracting multiples of 12. Similarly we have c4′ = u− c4 and c6′ = u− c6 and by the

same argument ν(c4) and ν(c6) can only be changed by adding or subtracting a multiple of 4 and 6, respectively. So we conclude:

1) If a R and ν(∆) < 12, then the equation is minimal. i ∈ 2) If a R and ν(c ) < 4 (or ν(c ) < 6), then the equation is minimal. i ∈ 4 6

Now we look at the “reduction modulo ̟”. Let us consider the natural reduction map R k = R/p. Let us denote bya ˜ k the reduction of a modulo ̟ and by E the −→ i ∈ i equation obtained from E by reducing its coefficients modulo ̟, that is e 2 3 2 E : y +ã1xy +ã3y = x +ã2x +ã4x +ã6.

The curve E is called thee reduction modulo ̟. Note that E/k is an elliptic curve if ∆˜ = 0, 6 this occurs when ν(∆) = 0. e e Deﬁnition 5.1.17. Let E/K be an elliptic curve and E its reduction modulo ̟. We say that e

a) E has good (or stable) reduction if E is non-singular (in which case E is an elliptic curve). e e b) E has multiplicative (or semi stable) reduction if E has a node. The reduction is called split if the tangent directions are deﬁned over k, otherwise is non-split. e c) E has additive (or unstable) reduction if E has a cusp.

e Twists

Definition 5.1.18. Let E and E1 be two elliptic curves defined over a field K. We say that E1 is a twist of E if E and E1 are isomorphic over an algebraic closure of K.

Remark 5.1.19. Two twists have the same j-invariant (cf. Theorem 5.1.4).

77 5. Applications and examples

Example 5.1.20. Let E be an elliptic curve defined over a finite field Fq given by the Weierstrass equation y2 = x3 + Ax + B.

Let β F×, then the elliptic curve ∈ q 2 3 2 3 E1 : y = x + β Ax + β B

4 6 is a F -twist of E. Indeed, taking c F 2 such that c = √β then (x, y) (c x, c y) is an q ∈ q 7−→ isomorphism from E to E1 deﬁned over Fq2 . This twist is called a quadratic twist.

5.2 Supersingular elliptic curves

Let K be a ﬁeld of characteristic p > 0 and E/K an elliptic curve. As mentioned in Section 5.1, the endomorphism ring of E is a torsion free Z-module of rank 1, 2 or 4. Namely, Z, an order in an imaginary quadratic ﬁeld or a maximal order in a quaternion algebra, respectively.

Deﬁnition 5.2.1. Let E/K be an elliptic curve, we say that E is supersingular if its endomorphism ring has rank 4.

There are further characterizations and interesting properties of supersingular elliptic curves (cf. [Sil09, Ch. III]).

Examples

1) If K is a ﬁeld of characteristic 2 then every elliptic curve with a Weierstrass equation

2 3 y + a3x = x + a4x + a6

is supersingular ([Was08, p. 122]).

2) If K is a ﬁeld of characteristic 3 then every elliptic curve of the form

2 3 y = x + a4x + a6

is supersingular ([Was08, p. 122]).

78 5.3. Elliptic curves over C and Eisenstein series

Supersingular elliptic curves will play an important role in our algorithm. We will need a test to know whether or not a given elliptic curve is supersingular.

Proposition 5.2.2. Let E be an elliptic curve over Fq, where q is a power of the prime number p and let a = q +1 #E(F ). Then E is supersingular if and only if a 0 − q ≡ (mod p).

For a proof of the proposition see for example [Was08, Prop. 4.31]. An important invariant in the theory of supersingular elliptic curves is the Hasse invariant which is deﬁned as follows.

2 Definition 5.2.3. Let E be an elliptic curve over Fq defined by the equation y = f(x), where f is a polynomial in Fp[x] of degree 3. The Hasse invariant of E is defined to be − p 1 (p 1) the coefficient of x − in the expansion of f(x) 2 .

Remark 5.2.4. In the literature it is common to deﬁne the Hasse invariant in terms of the diﬀerential associated to the elliptic curve (cf. [Kat77]).

Lemma 5.2.5. An elliptic curve E is supersingular if and only if its Hasse invariant is zero.

5.3 Elliptic curves over C and Eisenstein series

In this section we follow very closely the book of Silverman [Sil09, Ch. VI].

Deﬁnition 5.3.1. Let Λ C be a lattice, that is a discrete subgroup of C which contains ⊂ a R-basis for C. An elliptic function (relative to the lattice Λ) is a meromorphic function f(z) on C which satisﬁes

f(z + ω)= f(z) for all z C and ω Λ. ∈ ∈ The set of all elliptic functions for Λ forms a ﬁeld, denoted by C(Λ).

Deﬁnition 5.3.2. Let Λ C be a lattice. We deﬁne the Weierstrass ℘ -function (relative ⊂ Λ to Λ) as 1 1 1 ℘ (z) := + Λ z2 (z ω)2 − ω2 ω Λ,ω=0 ∈X6 −

79 5. Applications and examples

and also the Eisenstein series (for Λ) of weight 2k as

2k G2k(Λ) = ω− . ω=0 Λ X6 ∈

It is customary to let g2(Λ) = 60G4(Λ) and g3(Λ) = 140G6(Λ).

The Eisenstein series are absolutely convergent for all integers k > 2 and the series deﬁning

the Weierstrass ℘Λ-function converges absolutely and uniformly on every compact subset of C Λ (cf. [Sil09, Ch. VI, Thm. 3.1]). −

The ﬁeld C(Λ) is generated by the Weierstrass ℘Λ-function and its derivative. These functions can be used to parametrize certain elliptic curve as we see in the following theorem (cf. [Sil09, Ch. VI, Prop. 3.6]).

Theorem 5.3.3. Let Λ C be a lattice. ∈

a) The functions ℘Λ(z) and ℘Λ′ (z) generate C(Λ), that is, C(Λ) = C(℘Λ, ℘Λ′ ).

b) The Weierstrass ℘Λ-function and its derivative satisfy the identity

2 3 ℘′ (z) =4℘ (z) g (Λ)℘ (z) g (Λ). Λ Λ − 2 Λ − 3 Further, the polynomial f(x)=4x3 g (Λ) g (Λ) has distinct roots, so its discrim- − 2 − 3 inant ∆= g (Λ)3 27g (Λ)2 2 − 3 is non-zero and therefore the equation

E : y2 =4x3 g (Λ)x g (Λ) Λ − 2 − 3 deﬁnes an elliptic curve over C.

c) The map

φ : C/Λ E (C) z (℘ (z), ℘′ (z)) Λ −→ Λ 7→ Λ Λ is a complex analytic isomorphism of complex Lie groups.

d) Conversely, given an elliptic curve E/C, there exists a lattice Λ, unique up to homothety, such that EΛ ∼= E.

80 5.3. Elliptic curves over C and Eisenstein series

Let E1 and E2 be elliptic curves over C corresponding to lattices Λ1 and Λ2, respectively. Then one can show that

Hom(E , E ) = α C αΛ Λ , 1 2 ∼ { ∈ | 1 ⊂ 2} where the isogeny associated to α is given analytically by

C/Λ C/Λ , z αz. 1 −→ 2 7−→

We recall that two lattices Λ and Λ are homothetic if there exists α C such that 1 2 ∈ Λ1 = αΛ2. Homothetic lattices correspond to isomorphic elliptic curves, so it is common in practice to replace the lattice Λ := ω Z + ω Z by Λ := τZ + Z with τ = ω /ω 1 2 τ 1 2 ∈ H := τ C Im(τ) > 0 , which is homothetic to Λ. Then for the lattice Λ the { ∈ | } τ deﬁnition of the Eisenstein series becomes 1 G (τ) := G (Λ )= . 2k 2k τ (mτ + n)2k m,n Z (m,nX)=1∈

On the other hand, the fact that the lattice Λ := ω1Z + ω2Z does not change when we replace its basis ω ,ω by aω + bω ,cω + dω where a,b,c,d Z and ad bc = 1 { 1 2} { 1 2 1 2} ∈ − allows us to consider the modular group Γ = SL2(Z) and its action on H in order to study isomorphism classes of elliptic curves.

Deﬁnition 5.3.4. Let k be an integer. A holomorphic function f : H C is called a −→ modular form of weight k with respect to the modular group Γ = SL2(Z) if it satisﬁes the following conditions:

1) f aτ+b =(cτ + d)kf(τ) for all ( a b ) Γ, cτ+d c d ∈ 2) f has a Fourier expansion of the form

∞ n f(τ)= anq (5.7) n=0 X where q = e2πiτ .

A modular form with respect to Γ = SL2(Z) is a cusp form if it satisﬁes in (5.7) the further

condition that a0 = 0.

81 5. Applications and examples

a b 1 0 Remark 5.3.5. For ( c d )= −0 1 we deduce − f(τ)=( 1)kf(τ). −

So k must be even, otherwise f(τ)=0. In other words, there is no non-zero modular forms

with respect to Γ= SL2(Z) of odd weight.

A typical example of modular forms is given by the Eisenstein series Gk(τ) deﬁned above.

The coeﬃcients of Gk(τ) can be explicitly calculated as arithmetic functions on n. Namely, k 1 setting σk 1(n)= d n d − , we have the following proposition [Kob84, Ch. III, p. 110]. − | Proposition 5.3.6.P Let k be an even integer greater than 2 and let τ H. Then the ∈ modular form Gk(τ) has q-expansion

∞ 2k n Gk(τ)=2ζ(k) 1 σk 1(n)q − B − k n=1 ! X 2πiτ where q = e and the Bernoulli numbers Bk are deﬁned by

x ∞ xk = B . ex 1 k k! − Xk=0 It is convenient to deﬁne the normalized Eisenstein series

∞ 1 2k n Ek(τ)= Gk(τ)=1 σk 1(n)q . (5.8) 2ζ(k) − B − k n=1 X The series Ek(τ) is deﬁned in this way in order to have rational coeﬃcients. We have for example

∞ n E4(τ)=1+240 σ3(n)q , n=1 X ∞ E (τ)=1 504 σ (n)qn, 6 − 5 n=1 X ∞ n E8(τ)=1+480 σ7(n)q . n=1 X Let Mk(Γ) be the C-vector space of modular forms with respect to Γ = SL2(Z) of weight

k. We may use the standard notation Mk if the group Γ is clear from the context. It is clear that M M M , k l ⊂ k+l

82 5.3. Elliptic curves over C and Eisenstein series

and so the direct sum ∞ Mk Mk=0 can be viewed as a graded algebra, whose structure is given by the next theorem.

Theorem 5.3.7. Put Q = E4 and R = E6. The functions Q and R are algebraically independent and ∞ Mk = C[Q, R]. Mk=0 One has for example 441Q3 + 250R2 E = Q2, E = QR, E = , 8 10 12 691 Q3 R2 E = Q2R and ∆= − . 14 1728

Corollary 5.3.8. The dimension of Mk is given as follows:

k [ 12 ], if k 2 (mod 12) dim(Mk)= ≡ [ k ]+1, if k 0, 4, 6, 8, 10 (mod 12).  12 ≡

 2 The dimension of Mk is then 1 for k =0, 4, 6, 8, 10, with the basis 1, Q, R, Q , QR, respectively.

From Theorem 5.3.7 follows that for k even there exists a unique polynomial ϕ (X,Y ) k ∈ C[X,Y ] (actually in Q[X,Y ]) such that

ϕk(P, Q)= Ek. (5.9)

One has for example

ϕ4(X,Y )= X,

ϕ6(X,Y )= Y, 2 ϕ8(X,Y )= X ,

ϕ10(X,Y )= XY, 441X3 + 250Y 2 ϕ (X,Y )= , 12 691 2 ϕ14(X,Y )= X Y,

83 5. Applications and examples

X(1617X3 + 2000Y 2) ϕ (X,Y )= . 16 3617

5.4 Reduction modulo p of modular forms

In this section we closely closely the ideas introduced in [Ser73b]. Let p be a prime number

greater than 3 and let vp be the corresponding valuation of the ﬁeld Q. We can now deﬁne modular forms modulo p. Let us denote by o the local ring of Q at p, that is, the ring of rational numbers with denominator prime to p, and let m its maximal ideal.

Deﬁnition 5.4.1. Let f = a qn Q[[q]] n ∈ n>0 X be a formal power series with coeﬃcients in Q. We say that f is p-integral if vp(an) > 0 (equivalently if a o) for all n . n ∈

Leta ˜n denote the image of an in Fp = o/m. Let f be a p-integral power series, then

f˜ = a˜ qn F [[q]] n ∈ p X is called the reduction of f modulo p.

For a ﬁxed integer k, consider the following set

M := f˜ f is a modular form of weight k whose Fourier expansion is p-integral . k | n o f Denote by M the union of the Mk, which is a sub-algebra of Fp[[q]] and call it the algebra of modular forms modulo p. f f Deﬁnition 5.4.2. A polynomial is called isobaric if all monomials appearing in the polynomial have the same weight according to some given weight function on the indeterminates.

Let f be a modular form of weight k, we know from Theorem 5.3.7 that f may be written as an isobaric polynomial in Q and R, that is

a b f = ca,bQ R X for some ﬁnite set (a, b) such that 4a +6b = k, i.e., Q has weight 4 and R has weight 6.

84 5.4. Reduction modulo p of modular forms

Note that by equation (5.8) the denominators of the Eisenstein series are bounded, so we can normalize them to get integral coeﬃcients and therefore p-integral for any prime p. a b By Theorem 5.3.7 it follows that Mk admits a Fp-basis of monomials Q R , that is, Q and R generate the algebra M. To describe the structure of M we only need to determine the f e e e ideal a Fp[[X,Y ]] of relations between Q and R, i.e., those polynomials F for which e ⊂ f f F (Q, R) = 0. This is the content of the following (cf. [SD75, Thm. 2]) e e Theoreme e 5.4.3. Suppose that p> 3 is a prime. The ideal a is a principal ideal generated by A 1 where A F [X,Y ] is the isobaric polynomial of weight p 1 such that A(Q, R)= − ∈ p − Ep 1. The polynomial A(X,Y ) has no repeated factor and M is naturally isomorphic to − e e e F [X,Y ]/(A 1) f p − which has a natural Z/(p 1)-grading. −

Examples

For p = 5 one has Ep 1 = E4 = Q, so A(X,Y )= X. The ideal of relations among Q • − and R is generated by the relation Q = 1. The algebra M is isomorphic to F5[R]. e

For pe= 7 one has Ep 1 = E6 = R. Analogouse to the previousf case A(X,Y )= Yeand • − M = F7[Q].

Forf p = 11e one has E10 = QR, the fundamental relation is QR = 1, so that A(X,Y )= • XY . e e For p = 13 one has E 6Q3 5R2 (mod 12), the fundamental relation is • 12 ≡ − 6Q3 5R2 =1. − e e It is clear from Theorem 5.4.3 that A is a homogeneous polynomial of weight (p 1)/2, if − X and Y have weight 2 and 3, respectively.

Proposition 5.4.4. Let A be the polynomial of Theorem 5.4.3.

1. There exists a homogeneous polynomial F such that A(X,Y )= F (X3,Y 2)XiY j with i, j 0, 1 . ∈{ } 2. The exponents i and j depend on the congruence of p 1 modulo 12, namely −

85 5. Applications and examples

a) i = j =0 if and only if p 1 (mod 12). ≡ b) i =1 and j =0 if and only if p 5 (mod 12). ≡ c) i =0 and j =1 if and only if p 7 (mod 12). ≡ d) i =1 and j =1 if and only if p 11 (mod 12). ≡

Proof. Let A be the polynomial in Fp[X,Y ] that generates the algebra M then the fact that A(Q, R) = Ep and the graduation modulo p force A to be a pseudo-homogeneous polynomial of degree (p 1)/2 for X of weight 2 and Y of weight 3. Hencef we can write − A as e e e A(X,Y )= F (X3,Y 2)XiY j (5.10) for i, j N and F a homogeneous polynomial of degree d. Then pseudo-degree (ps-deg) of ∈ A is 6d +2i +3j where i 0, 1, 2 and j 0, 1 . We have six cases to consider. ∈{ } ∈{ }

1) If i = j = 0 then A(X,Y ) = F (X3,Y 2) and ps-deg(A)=6d, but ps-deg(A) = (p 1)/2 therefore 6d =(p 1)/2. That is p 1 (mod 12) and 2. a) follows. − − ≡ 2) If i = 1 and j = 0 we have A(X,Y )= F (X3,Y 2)X then ps-deg(A)=6d +2. From ps-deg(A)=(p 1)/2 it follows that 6d +2=(p 1)/2 and d =(p 5)/12, since d − − − is an integer, p 5 (mod 12) and we have the assertion of 2. b). ≡ 3) If i = 0 and j = 1 we have A(X,Y ) = F (X3,Y 2)Y then ps-deg(A)=6d + 3. In view of ps-deg(A)=(p 1)/2 we get d = (p 7)/12 and thus p 7 (mod 12). So − − ≡ we have the claim of 2.c).

4) If i = 1 and j = 1 we get A(X,Y ) = F (X3,Y 2)XY then ps-deg(A)=6d +5 as above, this implies that p 11 (mod 12) and 2.d) follows. ≡ 5) If i = 2 and j = 0 we have A(X,Y )= F (X3,Y 2)X2. Therefore ps-deg(A)=6d+4 = (p 1)/2 implies that d =(p 9)/12 that is 3 p which is impossible since p is prime. − − | 6) If i = 2 and j = 1 we have A(X,Y )= F (X3,Y 2)X2Y and then ps-deg(A)=6d +7 which implies 3 p and since p is prime, this case is impossible. |

86 5.5. The Tate Curve

Elliptic interpretation

Let p > 5 be a prime and E be an elliptic curve deﬁned over K := Fp(Q, R) with equation

Q R y2 =4x3 x (5.11) − 12 − 216

with Q and R regarded as indeterminates (compare the equation with (5.3)). As a result from the discussion in [SD75, pp. 21–24], the curve E has Hasse invariant H(E)= A(Q, R). Upon specializing (Q, R) the Hasse invariant H vanishes if and only if the corresponding e elliptic curve is supersingular. In [Gor02] the author proves even more.

Proposition 5.4.5. [Gor02, Prop. 5.3] The Hasse invariant is a modular form over Fp (of level) 1 and weight p 1. Its q-expansion at the cusp is 1, hence H is equal to the − reduction of Ep 1 modulo p. − Corollary 5.4.6. The Hasse invariant does not have multiple factors.

5.5 The Tate Curve

General references for the theory of Tate’s analytic uniformization of elliptic curves are [Lan87, Ch. 15] and [Sil94, Ch. V]. We refer to them for more details and for proofs of cited results.

In the classical case, for the ﬁeld of complex numbers, it is possible to represent the group of points on an elliptic curve over C as the quotient of the additive group of C by a discrete subgroup generated by two R-linearly independent periods ω1 and ω2. One can absorb one of these periods passing from the additive group to the multiplicative group, by means of the exponential function and obtain a representation of the group of points of the elliptic curve as the quotient of the multiplicative group C× by a discrete subgroup generated 2πiτ by one multiplicative period namely, t = e , where τ = ω2/ω1. In C, the explicit formulas giving this multiplicative representation are the well known Fourier expansions of the Weierstrass functions ℘, the Eisenstein series, etc.

Let K denote a ﬁeld which is complete with respect to a discrete valuation v, and whose residue ﬁeld is perfect of characteristic p> 0. Tate proved that the Fourier expansions of the Weierstrass functions ℘, suitable normalized, yield universal identities among power

87 5. Applications and examples

series that can be used to obtain a multiplicative representation for the group of rational points of certain elliptic curves over K (similar as in the classical case for the complex ﬁeld).

The following statements can be found in [Roq70, Ch. III, pp. 23–33].

Theorem 5.5.1 (Tate). Let K be a local field of arbitrary characteristic and let q K× ∈ with 0 < q < 1. Then the field of meromorphic q-periodic functions on G is an | | m,K elliptic function field F (q), which means, finitely generated of transcendence degree 1 over

K. More precisely F (q)= K(℘, ℘′) with

qnu q2nu2 ℘(u)= 2s and ℘′(u)= + s (1 qnu)2 − 1 (1 qnu)3 1 n Z n Z X∈ − X∈ − where

mkqm s = for k N. k 1 qm ∈ m>1 X − The elliptic curve E(q) associated to the elliptic function ﬁeld F (q) is given by the equation

2 3 ℘′ + ℘℘′ = ℘ + a4(q)℘ + a6(q), where a (q)= 5s and a (q)= 1 (5s +7s ). Its j-invariant is 4 − 3 6 12 3 5 (1 48a (q))3 1 j(q)= − 4 = + R(q) ∆ q

where

R(q) = 744+196884q + ... Z[[q]] and ∈ ∆(q)= a (q)2 a (q) 64a (q)2 + 72a (q)a (q) 432a (q)2. 4 − 6 − 4 4 6 − 6 To every j K with j > 1 there is one and only one q K with 0 < q < 1 such that ∈ | | ∈ | | j = j(q). The classical well known product representation

∆(q)= q (1 qn)24 − n 1 Y≥ holds also in the non-archimedean case for every characteristic.

88 5.5. The Tate Curve

The Tate elliptic curve (relative to q) is the curve with the Weierstrass equation

2 3 Eq : y + xy = x + a4(q)x + a6(q). (5.12)

7m5+5m3 qm 5 3 Observe that since a6(q) = m>1 12 1 qn and 7m 5m (mod 12) m Z, the − ≡ ∀ ∈ series a6(q) is also deﬁned forPp = 2 and p = 3.

Using the formulas above for the Weierstrass function and its derivative we obtain as in the classical case the v-adic analytic uniformization

∼= φ : K¯ × q Eq(K¯ ) h i −→ (5.13) u ℘(u), ℘′(u) . 7−→ 1 If char(K) = 2, 3, we can use the function ℘ + and its derivative ℘′ as the generators 6 12 of F (q) over K. It is immediately verified that their defining relation is in Weierstrass normal form 3 2 1 1 1 1 ℘′ = ℘ + g ℘ + g 12 − 4 2 12 − 4 3 where the coefficients g2 and g3 are given by the classical q-expansions 1 1 7 g = + 20s and g = + s . 2 12 3 3 −216 3 5

We call the corresponding elliptic curve

E : y2 =4x3 g x g , Eis − 2 − 3 the curve in Eisenstein form. It is clear that it is isomorphic to the Tate curve (5.12). It is a straightforward calculation to get the following relation between the coefficients of the Tate curve and the curve in Eisenstein form g 1 g g 1 a := a (q)= 2 and a := a (q)= 3 + 2 . 4 4 4 − 48 6 6 4 4 − 1728 Consider the Eisenstein form y2 = x3 g2 x g3 , then the Hasse invariant H gives a relation − 4 − 4 3 g2 g3 between the coefficients g2 and g3. Taking f(x)= x 4 x 4 we can calculate H as the (p−1) − − p 1 coefficient of x − in the expansion of f(x) 2 .

89 5. Applications and examples

One has for example

p = 5= H = g , ⇒ 5 2 p = 7= H = g , ⇒ 7 3 p =11= H = 11g g , ⇒ 11 2 3 p =13= H =8g2 +6g3, ⇒ 13 3 2 p =17= H =8g4 + 10g g2, ⇒ 17 2 2 3 p =19= H =7g3 + 11g3g2, ⇒ 19 3 2 3 p =23= H = 13g4g +9g g3, ⇒ 23 2 3 2 3 p =29= H =4g7 + 18g4g2 +8g g4, ⇒ 29 2 2 3 2 3 p =31= H =2g6g + g3g3 + 27g5. ⇒ 31 2 3 2 3 3

5.6 Obtaining the Tate parameter

Let N F [T ] be a polynomial of degree greater than 2 and ϕ Hnew(T, Z)Γ0(N) be a ∈ q ∈ ! harmonic cocycle with rational Hecke eigenvalues. From Section 2.12 we know that there exists an elliptic curve E deﬁned over F (T ) with conductor N associated with ϕ. ϕ q ∞ This curve E can be constructed as a Tate curve, that is there exists q F ((π)) with ϕ ∈ q 0 < q < 1 such that E is in the isogeny class of | | ϕ

2 3 Eq : y + xy = x + a4(q)x + a6(q).

From Proposition 2.12.2, we know that q is a generator of the multiplicative subgroup

c (α) α Γ (N) { ϕ | ∈ 0 }

where cϕ(α) is the multiplicative integral

t αz − 0 dµ (t) × t z ϕ Z∂Ω − 0 deﬁned in (3.9).

Now the ﬁrst question is how to ﬁnd α Γ such that c (α)= q. Let c , ...c be a basis ∈ ϕ { 1 g} for the homology of the quotient graph Γ T. Let ω be a path in the tree T without \ i

90 5.7. Obtaining equations for the curves

backtracking such that ωi is a lifting of the cycle ci. Since the origin v0,i and the terminal v of ω are Γ-equivalent, there exists a α Γ such that α v = v . Consider the set 1,i i i ∈ i 0,i 1,i C = α , ..., α with α as above, then from [Gek95, Cor. 3.19] we have that { 1 g} i val(q) = min val(c (α )) α C . (5.14) { ϕ i | i ∈ }

Therefore, to obtain the Tate parameter we choose an integral with minimal valuation. In the appendix we will describe a procedure to ﬁnd the minimal valuation without explicitly computing any integral. The algorithm to calculate the Tate parameter will be discussed in the Appendix B (cf. Algorithm 10), which can be calculated in polynomial time.

Theorem 5.6.1. The Tate parameter q can be calculated up to accuracy πM in time O(M 7).

We will probe this result in the appendix B (cf. Theorem B.2.1). This running time is strongly dominated by the running time for the algorithm to calculate the table. So after explain how calculate our table, we explain how to ﬁnd the Tate parameter and we give the proof of this theorem.

5.7 Obtaining equations for the curves

Let N F [T ] be a polynomial with deg(N) > 3 and such that the space of new harmonic ∈ q cocycles with rational Hecke-eigenvalues has dimension h. There exits h diﬀerent isogeny classes of elliptic curves deﬁned over F (T ) with conductor N . q ∞

new Γ0(N) For each one-dimensional rational eigenspace of H! (T, Z) we need to ﬁnd the Tate parameter associated to the corresponding rational eigencocycle ϕ. For this we use our algorithm to calculate the integral t αz − 0 dµ (t) × t z ϕ Z∂Ω − 0 for a suitable α obtained from (5.14).

Once we know the Tate parameter q up to accuracy πM for some ﬁxed integer M > 1, we

need to compute the quantities s3 and s5 using the formula

91 5. Applications and examples

mkqm s = = mkqm qml for k N. (5.15) k 1 qm ∈ m>1 m>1 > X − X Xl 0 Then we ﬁnd the coeﬃcients of the Tate curve given by a (q)= s and a (q)= 1 (5s + 4 − 3 6 12 3 M+1 7s5). We carry out these calculations modulo π since we only know q up to accuracy πM . Then the Tate equation is given by

2 3 Eq : y + xy = x + a4(q)x + a6(q). (5.16)

However, in general, the quantities a4(q) and a6(q) are not rational and this analytic model does not allow us directly to ﬁnd the isogeny class of the elliptic curve that we are looking

for. In this section, we explain how to get equations deﬁned over Fq(T ) by choosing suitable models for the elliptic curves.

5.7.1 Elliptic curves in characteristic 2 and 3

We need to transform the Tate curve in one model deﬁned over the rationals. For this we carry out in each characteristic an admissible change of variables to transform the Tate curve into a rational model depending only on one parameter.

In Characteristic 2

In this case we use the admissible change of variables

x x, 7−→ y y + x + a , 7−→ 4

then the Tate curve (5.16) is transformed into

2 3 E : y + xy = x + A6

2 where the coeﬃcients satisfy the relation A6 = a6 + a4. A direct calculation shows that E

has discriminant ∆ = A6 and j-invariant j =1/A6.

92 5.7. Obtaining equations for the curves

In Characteristic 3

An admissible change of variables in characteristic 3 is given by

x x + a , 7−→ 4 y y + a . 7−→ 4

Then the Tate curve (5.16) is isomorphic to

2 3 E : y + xy = x + A6

where A = a + a3 a2, which has discriminant ∆ = A and j -invariant j =1/A . 6 6 4 − 4 − 6 6 In both cases, since the j-invariant of such elliptic curve is a rational function, it follows

that A6 is rational. So the elliptic curve is deﬁned over Fq(T ).

The next proposition plays an important role in our algorithm (cf. [Sch01, Prop. 1.3]) in the cases p =2, 3.

Proposition 5.7.1. Suppose Fq is a ﬁnite ﬁeld of characteristic 2 or 3. Let E be an f(T ) elliptic curve over K = Fq(T ) with non-constant j-invariant j(E). Write j(E)= g(T ) with relatively prime f(T ) and g(T ) in Fq[T ]. Then

a) the divisors of f(T ) are places of supersingular reduction, and

b) the divisors of g(T ) are places of bad reduction.

Proof. For a) we know that in characteristic 2 and 3, an elliptic curve is supersingular if and only if the j-invariant is zero. So let p be a place that divides f(T ). Let E and ˜j the reduction of E and j modulo p, respectively. As p divides f(T ) we have that ˜j =0 so E is a supersingular curve, that is p is a supersingular place. e e Since j = 0 we can take the usual normal forms for characteristic 2 and 3 (cf. [Sil09, 6 Appendix A]) which have the advantage of describing an elliptic curve with an expression for the j-invariant relatively easy to manage.

An elliptic curve in characteristic 2 can be given by the Weierstrass model

2 3 2 E : y + xy = x + a2x + a6 (5.17)

93 5. Applications and examples

which has discriminant ∆ = a6 and j-invariant j =1/a6.

On the other hand in characteristic 3 an elliptic curve can be given by the normal model

2 3 2 E : y = x + a2x + a6 (5.18)

3 3 with discriminant ∆ = a2a6 and j-invariant j = a2/a6. Since the j-invariant of E is of f(T ) − − the form g(T ) we can write the equations of (5.17) and (5.18) as g(T ) E : y2 + xy = x3 + a x2 + 2 f(T ) and g(T ) E : y2 + xy = x3 + a x2 +2a3 , 2 2 f(T ) g(T ) respectively. The model in characteristic 2 has discriminant f(T ) and the one in character- 6 a2g(T ) istic 3 has discriminant f(T ) . Then for the proof of b), we can suppose without loss of generality that our elliptic curve has in each characteristic one of these models. From Tate’s algorithm [Tat75] we know that E has a model deﬁned over Fq[T ] which is minimal at all ﬁnite places. This model is also isomorphic to E and its discriminant is divisible by g(T ). Let p be a divisor of g(T ), then p divides the discriminant, that is valp(∆) > 0. Hence p is a bad place of E.

A prime p is a place of supersingular reduction if and only if the curve E is supersingular. By Proposition 5.2.2 E is supersingular if the corresponding Frobenious trace e t = q +1 #E(Fq) is 0 modulo p. But t is also an eigenvalue for the Hecke operator − e Tp. Then if we are able to calculate from the harmonic cocycle a list of eigenvalues for e primes of small degree, it is possible to have some candidates that divide the polynomial f(T ) of the proposition. On the other hand, what that polynomial g(T ) from last proposition concerns, places of bad reduction can be taken from the conductor which is also

known. So we can use Proposition 5.7.1 to improve our approximation of A6 to a rational function.

On the other hand, we know that val(j) = val(q), therefore we have that deg(g) − − deg(f)= val(q). Let us suppose that we have all the factors of f and g, so we can write − them as m n ti rj f(T )= fi and g(T )= gj (5.19) i=1 j=1 Y Y

94 5.7. Obtaining equations for the curves

for ti and rj integers. Then we have

n m r deg g t deg f = val(A ). j j − i i 6 j=1 i=1 X X

So we can try some m-tuples and n-tuples of ti’s and rj’s, respectively and take series

of A6 to get representation of A6 as a rational function. Although in our examples we do not need to do this and in practice this is diﬃcult to carry out since to calculate the supersingular places using the Hecke operator takes time.

At this point we have completed the ﬁrst phase of the computation with the approximation

of A6 to a rational function. Next we need to see whether the elliptic curve

2 3 y + xy = x + A6

has conductor N . The computation of the conductor is performed using any Computer ∞ algebra system with this function available, in our case we use MAGMA.

If the conductor of E is not N then we need to carry out an appropriated change of ∞ variables to transform our elliptic curve into a form that allows us to ﬁnd the right curve.

Remark 5.7.2. So far, in characteristic 2 this has not happened in any of the examples 2 3 known. That is, the model E : y +xy = x +A6 seems to give always the right conductor,so the algorithm assumes this.

In the following lines we describe the algorithm that allows us to ﬁnd a rational representative in the isogeny class of the Tate curve in characteristic 2 and 3 with conductor N . ∞ Given a rational Hecke eigen function ϕ Hnew(T, Z)Γ0(N) where N is a polynomial with ∈ ! deg(N) > 3, we want to ﬁnd the elliptic curve associated with ϕ. The Algorithm 1 explains the procedure to get E from the Tate period q.

Algorithm 1: Curve2or3 Input: The Tate parameter q up to accuracy of πM and the Hecke eigen-cocycle ϕ.

95 5. Applications and examples

Output: An elliptic curve in the isogeny class of the Tate curve associated to q or a message “Accuracy too small, please increase M”.

1: Calculate the quantities s3 and s5 using the formula (5.15) and use them to calculate M the coeﬃcients of the Tate curve a4 and a6 up to accuracy π .

2: calculate the quantity A6: if characteristic of K is 2 then A = a + a2, • 6 6 4 if characteristic of K is 3 then A = a + a3 a2. • 6 6 4 − 4 3: use the algorithm of continued fractions to ﬁnd the representation of A6 as a rational function. 4: for all place f divisor of N do

5: if f does not divide numerator of A6 then 6: Return “Accuracy too small, please increase M” 7: end if 8: end for 2 3 9: deﬁne the elliptic curve E : y + xy = x + A6 10: calculate the conductor of E let say c 11: if c = N then ∞ 12: Return E 13: end if 14: if characteristic K =3 then 2 15: for all places fi c and fi ∤ N, set u− =( fi) | fi 16: make the change of variables Q

x u2x, 7−→ y u3y + u2x 7−→

2 3 2 2 6 to transform E into En : y = x + a2x + a6 where a2 = u− and a6 = A6u− 17: end if 18: Return E

Remark 5.7.3. The Algorithm 1 ﬁnishes in step 6 if the accuracy M is too small and the series A6 does not converge to a rational function. In the other case it continues and

96 5.7. Obtaining equations for the curves

returns the curve E in step 12 if it is no necessary to carry out the change of variables or in step 18 after the change of variables.

If char K = 3 and j(E) = 0 we use the change of variables x u2x and y u3y + u2x to 6 7→ 7→ get the curve

E := y2 = x3 + a x2 + a ∆(E )= a3a , j(E )= a3/a n 2 6 n − 2 6 n − 2 6 2 for u K× such that u− is divisible by all extra places appearing in the conductor of E. We ∈ claim that E has conductor N . We cannot prove this, although we can give a heuristic n ∞ 12 argument as follows. From the change of variables we have that ∆(En) = u− ∆(E), and 2 a2 = u− (cf. [Sil09, Table 1.2]). Since j(E)=1/A6, by Proposition 5.7.1, the denominator

of A6 is divisible only by supersingular places and since we can not eliminate places of bad 12 reduction, we need u− to be the denominator of A6. We have observed in many examples

that the powers of the places that divide the denominator of A6 is 6.

12 Let Cond(En) be the conductor of En and p a place that divides u− , we have that

valp(Cond(En)) = 0 since p does not divide ∆(En). Then valp(Cond(En)) = valp(Cond(ETate)),

where ETate is the Tate curve.

Example 5.7.4. Let N = T 3 F [T ], from the quotient graph (see Figure 5.1), we can ∈ 2 see that the dimension of the space of harmonic cocycles is 1. Therefore, the dimension

new Γ0(N) of the space H! (T, Z) is 1 and consequently there exists one Fq(T )-isogeny class of elliptic curves with conductor T 3 . ∞ With a partition, as the one described in the previous chapter, using the Algorithm 6 we calculate with accuracy up to M = 85 to ﬁnd that

q = π4 + π36 + π68.

Then using the formulas for a4(q) and a6(q) we get

4 8 16 16 32 64 a4(q)= a6(q)= π + π + π + π + π + π .

2 From the change of variables we have the relation A6 = a6 + a4, so

8 16 16 32 64 4 8 16 16 32 64 A6 =(π + π + π + π + π )+(π + π + π + π + π + π ) = π4 =1/T 4.

97 5. Applications and examples

6 10 14 ❴❴❴ ?>=<89:; GFED@ABC GFED@ABC

?>=<89:;3 ❃ ❃❃ ❃❃ ❃❃ ❃❃ ❴❴❴ ?>=<89:;8 GFED@ABC12 GFED@ABC16

3 Figure 5.1: Quotient graph for N = T over F2.

2 3 1 The desired elliptic curve is given by the equation E : y +xy = x + T 4 which has conductor T 3 , and split multiplicative reduction at , as a routine application of Tate’s algorithm ∞ ∞ [Tat75] shows.

Remark 5.7.5. In [Pap01] Papikian ﬁnds the elliptic curve y2 + T xy = x3 + T 2x which is isomorphic to E after the change of variables

x T 2x, 7−→ y T 3y + T. 7−→

If q = 2, we can show using divisibility arguments and the deﬁnition of a4 and a6 that qn a = a = . 4 6 qn 6 1 n oddXn 1 − Also from the change of variables, we have the relation a2 + a = A where A F (T ). 4 6 6 6 ∈ 2 In other words, a is a root of the polynomial F (X) = X2 + X + A F (T )[X]. It is 4 6 ∈ 2 straightforward to check that if α is a root of F then 1 + α is the other root.

98 5.7. Obtaining equations for the curves

4 4 2 Let us consider the rational function A6 = 1/T = π and the polynomial F (X) = X + X + A F (T )[X]. A direct calculation shows that α = π2n is a root of F (X), 6 ∈ 2 n>2 hence, it is equal to a4 since the other root of F (X) has valuationP 0. On the other hand the continued fraction expansion of α is the inﬁnite periodic sequence

4 4 4 [1 + π− ; π− , π− , ...].

Then by the Lagrange theorem for continued fractions α / F (T ) and is quadratic over ∈ 2 F2(T ). In particular a4 and a6 are quadratic.

Once we know the rational function A6, we can use the formula for a4 and the polynomial F (X) = X2 + X + A F (T )[X] to arbitrarily increase the accuracy of q. To do 6 ∈ 2 M+1 this, suppose that one knows q up to accuracy M, then make q = q + bM+1π where b F is an indeterminate. Plug in q in the formula for a and use the polynomial M+1 ∈ 2 4 F (X) to ﬁnd the value of bM+1. With this procedure we can easily to compute q to an arbitrary high precision. For example:

q = π4 + π36 + π68 + π132 + π196 + π228 + π356 + π420 + π452 + O(π517).

Example 5.7.6. Consider N = T 3 over F . In this case dim H (Γ T) = dim H (T, Z)Γ0(N) = 3 1 \ ! 2 which is the number of new forms. In terms of the standard basis they are ϕ1 = (1, 1) and ϕ = (1, 2). For each of them we obtain, using the Algorithm 6 (see Appendix B), 2 − the following Tate parameters (working with an accuracy of M = 40)

3 12 21 39 q1 =2π +2π + π +2π , 3 12 21 39 q2 = π +2π +2π + π .

Then applying the formulas for a4 and a6 we have for q1 up to accuracy M = 40

3 9 27 a4(q1)=2π +2π +2π , 3 6 9 12 18 27 30 36 a6(q1)= π + π + π +2π + π + π +2π +2π and by the change of variables described above, we have A = a + a3 a2 = π3, so the 6 6 4 − 4 corresponding rational elliptic curve is

2 3 3 E1 : y + xy = x +1/T .

It has conductor T 3 and split multiplicative reduction at . ∞ ∞

99 5. Applications and examples

Analogously, for q2 we have

3 9 27 a4(q2)= π + π + π , 3 6 9 12 18 27 30 36 a6(q2)=2π + π +2π +2π + π +2π +2π +2π .

Performing the same calculations as above, we obtain the elliptic curve

2 3 3 E2 : y + xy = x +2/T .

Remark 5.7.7. In [Sch01, Prop. 4.3], Schweizer found exactly the same elliptic curves, using other tools.

Example 5.7.8. Consider now the polynomial N =(T +2)(T 2 +T +2). This example has a different flavor than the previous ones, since we do not get the curve immediately from the change of variables as above (in this case we have to apply step 16.b) of Algorithm 1). We need to make a further change of variables to find the curve with the right conductor.

In this case we have only one harmonic cocycle with rational eigenvalues, so it is also a new form. Working with accuracy of M = 40 and using our algorithm of integration, we ﬁnd the Tate parameter

q = π4 + π5 +2π7 +2π9 +2π10 +2π11 +2π14 + π15 +2π17 + π18 + π19 + π21 +2π22 +2π23 + π25 + π27 +2π28 +2π29 +2π32 + π33 +2π34 + π35 + π36 + π37 +2π38 +2π39 + π40.

Plugging in this value in the formulas for a4 and a6 we have that

4 5 7 9 10 11 14 15 16 18 A6 =2π +2π + π + π + π + π + π +2π +2π +2π + 2π19 +2π20 +2π23 + π24 + π25 + π27 + π28 + π29 + π32 +2π33 +2π34 +2π36 +2π37 +2π38.

Applying continuous fractions it converges to 2T 8 +2T 7 +2T 5 + T 4 + T 3 +2T 2 +2 A 6 ∼ T 12 +2T 9 + T 6 2(T + 2)4(T 2 + T + 2)2 = . T 6(T + 1)6 Then the elliptic curve 2(T + 2)4(T 2 + T + 2)2 E : y2 + xy = x3 + T 6(T + 1)6

100 5.7. Obtaining equations for the curves

has conductor T (T + 1)(T + 2)(T 2 + T +2) , which is not N . We need to “get rid of” ∞ ∞ the places T and T + 1. To do this, we change our model of the curve to the standard one in characteristic 3,

2 3 2 Enorm : y = x +ˆa2x +ˆa6.

So the admissible change of variables from E to Enorm is given by

x u2x, 7−→ y u3y + u2x. 7−→

2 3 With this change of variables we have that u aˆ2 = 1 anda ˆ6 = A6aˆ2. Also, from the table given in [Sil09], we have that the discriminants of E and Enorm satisfy the relation 12 1 u ∆E = ∆E . So to get rid of the places T and T + 1 we may take u = , norm √T (T +1) 2(T +2)4(T 2+T +2)2 thereforea ˆ2 = T (T +1) anda ˆ6 = T 3(T +1)3 . The new elliptic curve

2(T + 2)4(T 2 + T + 2)2 E : y2 = x3 + T (T + 1)x2 + norm T 3(T + 1)3

has the conductor (T + 2)(T 2 + T + 2) and split multiplicative reduction at . ∞

The following table gives some examples with the Tate parameter, the conductor N and

the rational function A6 in which we can see that the denominator has factors which are

powers of 6.

101 5. Applications and examples

Tate Parameter Conductor A6 2π5 + π6 + π7 +2π11 + π12 + π13 +2π20 + π21+ π22 +2π23 + π24 + π25 +2π29 + π30+ 31 32 33 34 35 36 37 2 T 5(T 2+2T +2) π + π +2π +2π + π +2π +2π + T (T + 2)(T +2T + 2) (T 2+1)6 π38 +2π39 +2π40 + O(π41)

π5 +2π7 + π8 +2π9 + π11 +2π13 + π14 +2π15+ π20 +2π22 +2π23 +2π24 +2π25 + π26 +2π27+ 29 31 32 33 34 36 37 2 (T +2)5(T 2+2T +2) π +2π +2π +2π +2π +2π + π + T (T + 2)(T +2T + 2) (T 2+T +2)6 π38 + π39 + π40 + O(π41)

π4 +2π5 + π7 +2π8 +2π13 + π14 + π16 +2π17+ 2π22 + π23 + π25 +2π26 +2π28 + π29 + O(π36) T 2(T + 2) T (T +2) (T +1)6

π4 +2π5 + π7 + π9 +2π10 + π11 +2π14+ 2π15 + π17 + π18 +2π19 +2π21 +2π22 + π23+ 25 27 28 29 32 33 2 2 (T +1)4(T 2+2T +2)2 2π +2π +2π + π +2π +2π + (T + 1) (T +2T + 2) T 6(T +2)6 2π34 +2π35 + O(π36)

π5 +2π7 + π8 +2π9 + π11 +2π13 + π14 +2π15+ π20 +2π22 +2π23 +2π24 +2π25 + π26 +2π27+ 29 31 32 33 34 36 2 (T +2)5(T 2+2T +2) π +2π +2π +2π +2π +2π + (T + 2)(T +2T + 2) (T 2+T +2)6 π37 + π38 + π39 + π40 + O(π41)

5.7.2 Elliptic curves over characteristic p> 3

In case char(K) =2, 3 it is more complicated to make a change of variables from the Tate 6 2 3 curve to a model depending only on one parameter as the curve y + xy = x + A6, used in characteristic 2 and 3. So we need to look for other model isomorphic to the Tate curve, in which we can claim rationality.

Let p be a prime greater than 3. Recall the deﬁnition of the Eisenstein series E4 = 1+240s (q) and E =1 504s (q) and deﬁne also their normalization as g = 1 E and 3 6 − 5 2 12 4 g = 1 E . 3 − 216 6 We consider now the following model for our elliptic curve

E : y2 =4x3 g x g (5.20) Eis − 2 − 3

102 5.7. Obtaining equations for the curves

with the admissible change of variables 1 x = x , − 12 1 1 y = y x + . − 2 24 We can transform the Tate curve into the Eisenstein curve where the coeﬃcients satisfy the known relations 1 g = a + − 2 4 48 1 1 g = a a + . − 3 6 − 12 4 864

np From Theorem 5.4.3 we now deduce that there exists integers np and mp such that g2 and mp g3 are rational and these integers depend on the class of p modulo 12 as follows.

Proposition 5.7.9. Let p> 3 be a prime number and g2 and g3 the normalized Eisenstein np mp series deﬁned as above. Then g2 ,g3 are in Fp(T ) where

p 1 p 1 ( − , − ) if p 1 (mod 12) 4 6 ≡ p 1 p 1 ( −4 , −2 ) if p 5 (mod 12) (np, mp)=  ≡  p 1 p 1 ( − , − ) if p 7 (mod 12)  2 6 ≡ p 1 p 1 ( −2 , −2 ) if p 11 (mod 12).  ≡   Proof. From Theorem 5.4.3 we have that there exists a polynomial A(X,Y ) F [X,Y ] ∈ p such that A(Q, R) = Ep 1 = 1 where Q and R are the reduction of the series E4 and E6 − modulo p, respectively. Since g2 and g3 are the normalized series of E4 and E6 it is enough to prove the claimse e foreQ and R. e e

From Proposition 5.4.4 we have that A is a modular form of weight (p 1)/2 and we can e e − write it as A(X,Y )= F (X3,Y 2)XiY j where F is a homogeneous polynomial of degree d and A has pseudo degree 6d +2i +3j = (p 1)/2. The exponents i and j depend on the congruence of p 1 modulo 12. − − 1 3 2 The elliptic curve (5.11) considered over Fp(Q, R) has discriminant ∆ = 1728 (Q R ) and 1728Q3 Q3 − j-invariant j = e . Since j is a rational function it is easy to see that e is a rational ∆ R2 e e Q3 e e f function, that is, there exists a f F (T ) such that e = f. p R2 ∈ e

103 5. Applications and examples

Let us write d 3l 2(d l) i j A(X,Y )= X X − X Y αl (5.21) Xl=0 for suitable α F . In view of A(Q, R) = 1 and Q3 = fR2 we have that l ∈ p e e d e e 3l 2(d l) i j A(Q, R)= Q R − αl Q R (5.22) ! Xl=0 e e d e e e e l 2l 2d+j 2l i = f R R R− Q αl Xl=0 d e e e e l 2d+j i = f R Q αl Xl=0 d e e l 2d+j i = f αl R Q ! Xl=0 =1. e e

On the other hand, we have also that

d 3l 2(d l) i j A(Q, R)= Q R − αl Q R (5.23) ! Xl=0 e e d e e e e 3l 3d+i (d l) 3l j = Q Q f − − Q− R αl Xl=0 d e e e e (d l) 3d+i j = f − − Q R αl Xl=0 d e e (d l) 3d+i j = f − − αl Q R ! Xl=0 =1. e e

Then from (5.22) and (5.23) we have that R2d+jQi and Q3d+iRj are rational functions for i, j 0, 1 . ∈{ } e e e e

1. If p 1 (mod 12) then from Proposition 5.4.4 A(X,Y ) = F (X3,Y 2) that is, i = ≡ j = 0 then A has pseudo degree 6d = (p 1)/2 and we have that R2d and Q3d are (p 1)/6 (p−1)/4 rational functions, hence R − and Q − are in Fp(T ). e e e e

104 5.7. Obtaining equations for the curves

2. If p 5 (mod 12) we have in this case that A(X,Y ) = F (X3,Y 2)X with pseudo- ≡ degree 6d +2=(p 1)/2. Since R2d+jQi and Q3d+iRj are rational functions then − for i = 1 and j = 0 we get R2dQ, Q3d+1 F (T ). Since 3d +1= (p 1)/4, ∈ p − (p 1)/4 e e6d 3 e e 3 2 Q − Fp(T ). On the other hand, R Q is also rational, replacing Q by fR ∈ 6d+2 e e6d+2e (p 1)/2 we get that R f Fp(T ) then R = R − is rational. e ∈ e f e e 3. If p 7 (mode 12) then A(X,Y ) =e F (X3,Ye 2)Y . In this case A has pseudo-degree ≡ 6d+3=(p 1)/2. We can proceed as above replacing i and j by 0 and 1, respectively. − 2 2d+1 (p 1)/6 3d 6d 2 3d We have that R = R − and Q R are rationals. Also Q R = Q R is 2 3 1 6d+3 1 rational. Then since R = Q f − we have that Q f − Fp(T ) which implies that 6d+3 (p 1)/e2 e e e ∈ e e e e Q = Q − Fp(T ). ∈ e e e 4. Ife p 11e (mod 12) then the polynomial A is A(X,Y )= F (X3,Y 2)XY with pseudo- ≡ 2d+1 3d+1 degree 6d +5=(p 1)/2 and therefore R Q and Q R are in Fp(T ). In view of 3 2 − 6d+3 2 6d+5 (p 1)/2 and Q = fR we have that R R f is rational. Then R = R − Fp(T ). 6d+2 3 1 e e e e(p 1)/2 ∈ Analogously, Q Q f − is rational, which implies that Q − Fp(T ). e e e e e ∈e e e e

np f(T ) mp r(T ) Let us suppose that we have g2 = g(T ) and g3 = s(T ) for some polynomials f(T ),g(T ),r(T ), s(T ) F [T ] (no necessarily monic cf. Example 5.7.15). Then we have that the curve E ∈ p Eis (5.20) is

f(T ) 1/np r(T ) 1/mp y2 =4x3 χ x ξ − np g(T ) − mp s(T )

where χnp and ξmp are np-th and mp-th roots of the unity in Fp, respectively, note that n , m p 1, so that χ and χ indeed lie in F . So the curve E is deﬁned over a ﬁnite p p| − np np p Eis extension of Fp(T ).

Let E be an elliptic curve over a ﬁeld K of the form Y 2 = X3 + AX + C and u K then ∈ 2 3 2 3 the elliptic curve Eu : Y = X + u AX + u B is a twist of E. Taking u =1/√A the curve 2 3 B Eu is Y = X + X + 2 . √A3

2 3 When E is EEis : y =4x g2x g3 we can carry out the twist as follows, let us suppose − − 1/np np mp 2 g(T ) that g2 and g3 are as above, then taking u = f(T ) we get

105 5. Applications and examples

f(T ) 1/np r(T ) 1/mp y2 =4x3 u2ξ x u3χ − np g(T ) − mp s(T ) r(T ) 1/mp g(T ) 3/2np y2 =4x3 ξ x χ . − np − mp s(T ) f(T )

Remark 5.7.10. This twisted elliptic curve may or not be deﬁned over Fp(T ), because the 1/m 3/2n r(T ) p g(T ) p quantity s(T ) f(T ) is not always rational. However, by reviewing examples, heuristically we can say that after simpliﬁcations this expression becomes F (T )G(T )1/2 for some rational functions F (T ) and G(T ). If G(T ) is not one, then we need to make another 1/2 twist by taking u =(G(T ))− . The resulting elliptic curve has always the right conductor.

106 5.7. Obtaining equations for the curves

Algorithm 2: Tateforp> 3 Input: The Tate parameter q up to accuracy of πM . Output: An elliptic curve in the isogeny class of the Tate curve associated to q or a message “Accuracy too small, please increase M”.

1: Calculate the quantities s3 and s5 using the formula (5.15) and use them to calculate M the coeﬃcients of the Eisenstein curve g2 and g3 up to accuracy π .

2: set the quantities np and mp as the exponents of g2 and g3, respectively. According to Proposition 5.7.9 np mp 3: use the algorithm of continued fractions to ﬁnd the representation of g2 and g3 as a np f(T ) mp r(T ) rational functions. Let say g2 = g(T ) and g3 = s(T ) .

4: deﬁne two list Lnp and Lmp which contain the np-th and mp-th roots of the unity in

Fp, respectively. 5: for ξ L do np ∈ np 6: for χ L do mp ∈ mp 7: deﬁne the elliptic curve

r(T ) 1/mp g(T ) 3/2np E : y2 =4x3 ξ x χ − np − mp s(T ) f(T ) . 1/mp 3/2np r(T ) g(T ) 1/2 8: write s(T ) f(T ) as F (T )G(T ) . ⊲ cf. Remark 5.7.10. 9: if G(T) =1 then 6 1/2 10: set u =(G(T ))− and deﬁne the twisted curve

2 3 1 1 E : y =4x ξ G(T )− x χ F (T )G(T )− − np − mp 11: end if 12: end for 13: end for 14: calculate the conductor of E let say c 15: if c = N then ∞ 16: Return E 17: else 18: Return “Accuracy too small, please increase M ” ⊲ This message is printed if after considering all possible elliptic curves we do not get the right conductor.

107 5. Applications and examples

19: end if

Remark 5.7.11. If the prime p is a large prime then the numbers np and mp are also big then one needs high accuracy to get convergence, in this cases it is better to consider other twist. Let u = g2 then the twisted curve is g3 g3 g3 2 3 2 3 2 3 2 2 y =4x u g2x u g3 y =4x 2 x 2 − − − g3 − g3

3 g2 2 since 2 is a rational function, the curve is deﬁned over K. The j-invariant of y = g3 3 A 4x Ax A is j = 1728 A 27 . Solving for A shows that j K if and only if A K. − − − ∈ ∈ However the cost of this twist is that a lot of extra places may appear in the conductor, which one has later to get rid of. One can see, for instance Example 5.7.12 that the present form is not isogenous over K to the searched Weierstrass equation. It is only correct up to some unspeciﬁed quadratic twist.

Example 5.7.12. In characteristic 5.

With p = 5 consider the polynomial N = T 2(T 1). The dimension of the space of − harmonic cocycles is 4, then we have 4 isogeny classes. Let ϕ one of these harmonic

cocycles with rational Hecke eigenvalues. In characteristic 5, we have that E4 = 1 and

g2 = 3 and the Eisenstein form is

E : y2 =4x3 +2x g . Eis − 3

(p 1)/2 Since p 1 is divisible by 4 and not by 6 then we have that g − F (T ). With an − 3 ∈ 5 accuracy of M = 70 we get the Tate parameter

q = π2 +2π4 + π5 +4π6 +3π8 + π9 +2π10 +4π11 +4π12 +2π13 + 4π14 +3π15 + π16 +2π17 +3π19 +4π20 + π21 +2π22 +2π23 +3π24 + 4π26 +3π27 +2π28 +2π29 +2π30 +2π31 +4π33 +3π34 + π35 +2π36 + 4π37 + π39 +3π40 +2π41 +2π42 + π43 +2π44 +4π45 +3π46 +4π47 + 4π48 +2π50 + π51 +3π52 + π53 +3π55 +4π56 +3π57 +2π58 +2π59 + 2π60 +2π61 +4π62 + π64 +3π65 +2π66 +3π67 +4π68 +4π69 +4π70 + O(π71).

Plugging in this value in the equation of g3 and applying the continuous fraction method 2 T (T +2)2 we have that the series g3 converges to the rational function (T +3)3 . Using Algorithm 2

108 5.7. Obtaining equations for the curves

T +2 T with F (T )= T +3 and G(T )= T +3 , the equation for the elliptic curve EEis is

2 3 2 3 T +2 T y =4x +2x g3 y = x +2x ξ2 − − T +3rT +3 3T +4 4T +3 y2 = x3 + x + T T

which has conductor T 2(T 1) . To get the last equation we make a quadratic twist with − ∞ T +3 u = T and ξ2 = 4. q 2 3 2 2 3 Remark 5.7.13. In [Sch11] is given the elliptic curve E1 : y = x +4T x +4T x as a representative of the isogeny class of EEis. They are isomorphic, namely the isomorphism is given by

E E Eis −→ 1 (x, y) (T 2x +2T 2, T 3y). 7→

Example 5.7.14. In characteristic 7.

Let us consider the polynomial N = T 3 2 over F . In this case we have that the dimension − 7 of the space of harmonic cocycles is 1 so there is only isogeny class. With an accuracy of M = 60 we ﬁnd the value of the Tate parameter is

q =5π3 +4π6 + p9 +6π12 +2π15 + p18 + p21 +2π24 +p27 +3π30 +5π33 +2π36 +5π39 +5π42 +5π45 + 4π48 +5π51 +6π54 +4π57 +5π60 + O(π61).

When p = 7 we have that g3 = 6 and also from the considerations of the modular forms 3 modulo 7, we have that g2 is a rational function. With the value of q and the formula for g2 we have that

3 6 9 12 15 18 21 24 g2 =1+3π +3π + π +2π +5π +3π +4π +5π + 4π27 + π30 +2π33 +5π36 +3π39 +4π42 +5π45 + 4π48 + π51 +2π54 +5π57 +3π60 + O(π61)

3 and g2 converges to (T 3 + 2)3T 3 . (T 6 +2T 3 + 2)2

109 5. Applications and examples

Then our model for the elliptic curve in the isogeny class is

3 3 3 2 3 2 3 3 (T + 2) T y =4x g2x +1 y = x + ξ3 x +5 − s(T 6 +2T 3 + 2)2 y2 = x3 + T (T 3 + 2)x + 5(T 6 +2T 3 + 3).

6 3 3 T +2T +2 We get the last equation by twisting with u = (T 3+2)3T 3 and using ξ3 = 1. q Example 5.7.15. In characteristic 11.

2 Consider the polynomial N = (T + 8)(T + 9) deﬁned over F11, in this case there are 4 diﬀerent isogeny classes, taking one the harmonic cocycles with rational Hecke eigenvalues and with accuracy M = 40 we have the following Tate parameter

q =4π +6π2 +4π3 +2π4 +4π5 + 10π6 +9π8 +9π9 + 10π10 +8π11 + 7π12 + π13 +7π14 + 10π15 + 10π16 +2π17 +9π18 +6π19 +9π20 +6π21 + 6π22 +7π24 +8π25 +3π26 +2π27 +4π28 + π29 +4π30 + π31 +4π32 + 4π33 +9π34 +9π35 +8π36 +6π37 +3π38 + 10π39 +4π40 + O(π41).

5 5 Using the relations of Serre and Swinnerton-Dyer, we have that g2 and g3 are rationals.

Plugging in the value of q in the formulas for g2 and g3 we have that

5 2 3 4 5 6 7 9 10 g2 =1+4π +9π +8π +8π +2π +9π +7π +9π +6π + 3π11 +5π12 +3π13 + 10π14 + 10π15 +8π16 +3π17 +6π18 + 3π20 +2π21 + π22 +9π23 + π24 +7π25 +7π26 + 10π27 + π28 + 2π29 + π31 +8π32 +4π33 +3π34 +4π35 +6π36 +6π37 +7π38 + 4π39 +8π40 + O(π41) and

5 2 4 5 6 7 8 9 10 g3 =10+4π +4π +6π +2π +6π +5π +7π +8π +4π + 6π11 +8π12 +8π13 + π15 +4π16 + π17 + 10π18 +3π19 +5π20 + 8π21 + π22 +5π23 +5π24 +2π26 +8π27 +2π28 +9π29 +6π30 + 10π31 +5π32 +2π33 + 10π34 + 10π35 +4π37 +5π38 +4π39 +7π40 + O(π41),

T 2+7T +4 (T +9)2 10T 2+8T +6 10(T +7)2 2 which converge to T 2+3T +5 = (T +7)2 and T 2+7T +4 = (T +9)2 , respectively. Taking u =

110 5.7. Obtaining equations for the curves

5 T +7 2 √5 T +9 , ξ5 = 9 and 10 = 2, the elliptic curve we are looking for is q 2 2 2 3 2 3 5 (T + 9) 5 10(T + 7) y =4x g2x g3 = y = x ξ5 x − − ⇒ − s(T + 7)2 − s (T + 9)2

T +7 2 T +7 3 = y2 = x3 ξ x 2 5 5 ⇒ − 5 − T +9 T +9 s s

T +7 = y2 = x3 +6x + ⇒ T +9 which has conductor (T + 8)(T + 9)2 . ∞ The other three isogeny classes are given by

2 3 4 2 2 4 3 E2 : y = x + 6(T +6T +6T + 8)x + 10(T +2T + 6)(T + 12T +6T + 9), 2 3 2 2 2 2 E3 : y = x + 9(T + 11T + 9)x +(T + 12) (T +4T + 2), 2 3 2 2 E4 : y = x +(T + 4) x +(T + 2)(T + 6)(T + 10).

Example 5.7.16. In characteristic 13.

3 Consider the polynomial N = T + 11 over F13. For this case there is only one harmonic cocycle with rational Hecke eigenvalues and with accuracy of M = 40 we have the following Tate parameter

q =2π3 +8π6 +9π9 + π12 +6π15 + 12π18 +9π21 +4π24 +3π27 + 2π30 +7π33 +7π36 +7π39 + O(π41).

3 2 In this case we have that g2 and g3 are in F13(T ), then using the value of q and the formulas for g2 and g3 we have that

3 3 6 9 12 15 18 21 24 g2 =1+10π +2π +6π +6π + π +7π + 11π +3π + 7π27 + π30 + 12π36 + O(π41) and

2 3 6 9 12 15 18 21 24 27 g3 =1+12π +5π +2π +2π +9π + 11π +8π + π + 11π + 9π30 +4π36 + O(π41).

111 5. Applications and examples

Using continuous fractions we see that

T 3(T 3 + 3)3 g3 2 (T 6 +2T 3 + 9)(T 6 + 10T + 6)

and (T 3 + 2)2(T 3 + 10)2 g2 . 3 (T 6 +2T 3 + 9)(T 6 + 10T + 6)

3 2 Using these values for g2 and g3, proceeding as in the examples above with 6 3 6 2 3 (T +2T +9)(T +10T +6) u = T 3(T 3+3)3 we have that q

2 3 2 3 3 3 2 1 y =4x g2x g3 = y = x ξ3x (T + 2)(T + 10) − − ⇒ − − sT (T 3 + 3)

= y2 = x3 +3T (T + 3)x + 3(T 3 + 2)(T 3 + 10). ⇒ We get the equation using the twist described in Algorithm 2 with F (T )=(T 3 +2)(T 3 +10) 1 and G(T )= T (T 3+3) .

112 A. Algorithms for the Quotient graph

In this appendix we shall provide the reader with some of the algorithms used in the thesis to deal with quotient graphs. In 2 we give a set of representatives for the edges of the § quotient tree Γ T where Γ = GL (A), and also give a routine, which we call decom, to 0 \ 0 2 calculate classes in Γ0 GL2(K )/I . In 3 we use the algorithm decom to lift a cycle in \ ∞ ∞ § the quotient graph Γ (N) T to a path in T. The algorithm decom gives a routine to solve 0 \ the following problem: given two matrices g1,g2 GL2(K ) which are in the same class ∈ ∞ in Γ0(N) GL2(K )/I , there exist γ Γ0(N) and κ I such that g1 = γg2κ. The \ ∞ ∞ ∈ ∈ ∞ algorithm that allows us to write such as decomposition is given in 4. § Also, we include a section that sets out the key deﬁnitions and results of computer arithmetic, which we need to give the running time of the main algorithms that allows us to calculate the Tate parameter.

A.1 Computational complexity of mathematical operations

The aim of this section is to give a brief summary of some fundamental definitions and results concerning computer arithmetic, algorithms for arithmetic in finite fields and polynomial rings. The intention is not to provide an implementation guide, instead, we state some complexity results that will be used later in the appendices of this thesis. More details of these subjects can be found in [vzGG13].

Since computers do not work on numbers but with data, so the very ﬁrst issue is how to feed numbers as a data into a computer. Data are stored in pieces called words. Current machines use either 32 or 64-bit words; in this thesis we assume that we have a 64-bit processor. Integers are represented as a sequence of binary words. We think of an algorithm

113 A. Algorithms for the Quotient graph as a sequence of word operations. The analysis of running time of an algorithm (or just running time) quantiﬁes the amount of time taken by the algorithm to run as a function of the length of the string representing the input. We can think of the running time as the number of statements executed by the program or as the length of time taken to run the program on some standard computer. It is common to estimate their complexity in the asymptotic sense, i.e., estimating the complexity function for arbitrarily large input, so we introduce the following deﬁnition.

Deﬁnition A.1.1. Let f and g be two functions deﬁned on some subset of the real numbers. One writes f(x)= O(g(x)) as x →∞ if and only if there exists a positive real number C and a real number x0 such that

f(x) C g(x) for all x x . | | ≤ | | ≥ 0

Let R be a commutative ring with 1. Operations like add or multiply may correspond to many bit or word operations. As a general rule, we will consider the number of arithmetic operations (additions and multiplications) in the ring R, (divisions, if R is a ﬁeld) used by an algorithm. The other operations such as index calculations or memory accesses, tend to be of the same order of magnitude. These are usually performed with machine instructions on single words (for example move a pointer in an array, etc), and their cost becomes negligible when the arithmetic quantities are large. The next result can be found in [vzGG13, Cor. 4.7].

Lemma A.1.2. Let q = pn with p a prime number and n > 1. One arithmetic opera- 2 tion, that is, addition, multiplication, or division, over Fq can be done using O(n ) word operations, where n = log (q)/64 +1. ⌊ 2 ⌋

Let R[x] be the polynomial ring with coeﬃcients in R. The basic algorithms for addition, subtraction, multiplication, and division of polynomials are quite straightforward adapta- tions of the corresponding algorithms for integers. In fact, because of the lack of “carry overs” these algorithms are actually much simpler in the polynomial case. We have the following easy result.

Lemma A.1.3. Let f and g be arbitrary polynomials in R[x] of degree n and m, respectively.

114 A.2. Representatives for the edges of Γ T 0 \

1. We can compute f + g with O(n + m) operations in R.

2. We can compute fg with O(nm) operations in R.

3. If g = 0 and the leading coeﬃcient of g is a unit in R, we can compute q,r R[x] 6 ∈ such that f = gq + r and deg(r) < deg(g) with O(deg(g) deg(q)) operations in R.

Remark A.1.4. Throughout the book [vzGG13], the authors discuss four algorithms to do fast multiplication of polynomials, the classical one, Karatsuba ([ Ibid., 8.1]), Fast § Fourier Transform (FFT) ([ Ibid., 8.2]) and Sh¨onhage & Strassen ([ Ibid., 8.3]. Due to § § the variety of multiplication algorithms, we introduce the following deﬁnition.

Deﬁnition A.1.5. Let R be a commutative ring with 1. We call a function M : N R −→ >0 a multiplication time for R[x] if polynomials in R[x] of degree lest than n can be multiplied using at most M(n) operations in R.

The following table summarizes the multiplication times for the algorithms mentioned above.

Algorithm M(n) Classical 2n2 Karatsuba O(n1.59) FFT O(n log(n)) Sh¨onhage & Strassen O(n log(n) log(log(n)))

A.2 Representatives for the edges of Γ T 0 \

In the first part of this section, we recall some definitions, in order to fix the notation. We define the Iwahori subgroup of GL2(K ) as ∞ a b I := GL2(O ) c 0 (mod π) . (A.1) c d ∈ ∞ ≡

We denote by I = IK× . Analogously, K and K denote the groups GL2(O ) and ∞ ∞ ∞ ∞ GL2(O )K× , respectively. ∞ ∞

115 A. Algorithms for the Quotient graph

In the rest of this appendix, we denote by Γ0 the group GL2(A) andby Γ0(N) the subgroup of Γ0 consisting of matrices which are upper triangular modulo N, where N is a polynomial in A = Fq[T ].

The Bruhat-Tits tree for GL2(K ) is denoted by T (cf. 2.4), while the quotients Γ0 T ∞ § \ and Γ (N) T are denoted by G and G , respectively. We use also the notation G for 0 \ 1 N N,M the quotient graph GN with the cusps up to level M.

The classes in GL2(K )/I will be denoted by [ ]1 and the classes in GL2(K )/K by ∞ ∞ · ∞ ∞ [ ] . Also the classes of edges and vertices in the double quotient Γ T are denoted by J K · 0 0 \ · 1 and J K , respectively. In the case of the quotient graph by the subgroup Γ (N) we add · 0 0 the subindex N.

The edges and vertices of a graph G will be denoted by Y (G) and X(G), respectively.

Notation: Let H be a subgroup of a group G. We say that S = s ,s , ... G is a set { 1 2 } ⊆ of representatives for the quotient H G if for all g G, there exist h H and a unique \ ∈ ∈ s S such that g = hs . Analogously for H and H subgroups of G, we say that S is a set i ∈ i 1 2 of representatives for the double quotient H G/H if for all g G, there exist h H , 1 \ 2 ∈ 1 ∈ 1 h H and a unique s S such that g = h s h . 2 ∈ 2 i ∈ 1 i 2

We know that G is isomorphic to the subgraph in T whose vertices are [k, 0] > (cf. 1 { }k 0 Lemma 2.4.4 for the deﬁnition). The following theorem (cf. [Ser03, p. 87]) gives a set of representatives for the vertices of G1.

Proposition A.2.1. A set of representatives for Γ0 GL2(K )/K is given by \ ∞ ∞ 1 0 > Λn = n for n 0 . ( 0 π ! )

That is, given g GL2(K ) there exist a unique n > 0, γ Γ0 and α K such that ∈ ∞ ∈ ∈ ∞

g = γΛnα.

Notation: Let g GL2(K ), we say that the vertex represented by g has “level” n if ∈ ∞ g JΛ K . ∈ n 0 In [But07, Lemma 18] one can ﬁnd a constructive proof of Proposition A.2.1. There, the author explains how to decompose any g GL2(K ) as g = γΛnα. So in this appendix ∈ ∞ we assume that we already have an algorithm that allows us to write such decomposition.

116 A.2. Representatives for the edges of Γ T 0 \

Before giving a set of representatives for the edges of G1 we need to define some functions and sets that will be useful in this section. Let us define first the function origin as follows

o1 := Γ0 GL2(K )/I Γ0 GL2(K )/K . (A.2) \ ∞ ∞ −→ \ ∞ ∞ JgK o (JgK )= JgK 1 7−→ 1 1 0

That is, given g GL2(K ) we deﬁne the origin of the class Γ0gI to be the vertex ∈ ∞ ∞ represented by Γ0gK . The function o1 is surjective and is not injective. So it is important ∞ to describe the ﬁbers of classes in Γ0 GL2(K )/K . \ ∞ ∞

Let g GL2(K ) then ∈ ∞

1 o1− (Γ0gK )= Γ0hI Γ0hK =Γ0gK . ∞ { ∞ | ∞ ∞}

Let g1,g2 GL2(K ) such that Γ0g1K = Γ0g2K and Γ0g1I = Γ0g2I . That is, g1 ∈ ∞ ∞ ∞ ∞ 6 ∞ and g2 represent diﬀerent edges with the same origin, then g1 Γ0g2K and g1 / Γ0g2I . ∞ ∞ ∈ 1 ∈ Hence in order to understand how many classes there are in o1− (Γ0gK ) we need to ∞ study the number of orbits of Γ0 in g2K /I , where Γ0 acts g2K /I via left matrix ∞ ∞ ∞ ∞ multiplication.

A straightforward calculation shows that there is a 1-1 correspondence between K /I = ∞ ∞ 1 K/I and P (Fq) given by

K/I P1(F ). −→ q a ( c ∗ ) I (a (mod π): c (mod π)). ∗ 7−→ Hence a set of representatives for K/I is

s 1 1 0 R = s F . 1 0 ∈ q 0 1 [ Proposition A.2.2. Let g GL2(K ). ∈ ∞

1. If g = I2 then #Γ0 K /I =1. That is, Γ0 acts transitively on K /I . \ ∞ ∞ ∞ ∞

2. If g / JI2K1 then #Γ0 gK /I = 2 and Γ0 gK /I = Γ0gI , Γ0gs1I where ∈ \ ∞ ∞ \ ∞ ∞ { ∞ ∞} 0 1 s1 =( 1 0 ) .

117 A. Algorithms for the Quotient graph

Proof. 1. Let ( s 1 ) be an element of R, for a α β Γ we have 1 0 γ δ ∈ 0 α β s 1 αs + β α = . (A.3) γ δ ! 1 0 ! γs + δ γ !

1 0 0 1 From (A.3) we see that I2 = ( 0 1 ) is an element of this orbit since ( 1 s ) Γ0 and − 0 1 s 1 1 0 s 1 ∈ ( 1 s )( 1 0 )=( 0 1 ). Then the orbit of I2 intersect all orbits of ( 1 0 ) for all s Fq. − ∈ Hence there exists only one orbit and #Γ0 K /I = 1. \ ∞ ∞ 2. If g / JI K, without loss of generality (Prop. A.2.1) we may consider g = π−n 0 for ∈ 2 0 1 > some n 1, that is,

n 1 0 π− 0 g = n n 0 π ! 0 π− ! n π− 0 =Λn n . 0 π− !

s 1 s′ 1 First note that for s,s′ Fq with s = s′ the orbits of ( 1 0 ) and of ( 1 0 ) intersect ∈ 1 T n(s′ s) 6 each other, indeed, let γ = − Γ , then γ veriﬁes 0 1 ∈ 0 n n n n n 1 T (s′ s) sT T s′T T − = . 0 1 ! 1 0 ! 1 0 ! On the other hand, for all s F the orbit of I does not intersect the orbit of ( s 1 ). ∈ q 2 1 0 (n) (n) 0 1 It is enough to prove that Γ0g0 I =Γ0g0 ( 1 0 ) I . Suppose to the contrary that ∞ 6 ∞ they are equal, then we have

(n) (n) (n)−1 (n) Γ0g0 I =Γ0g0 s1I s1 g0 Γ0g0 I ∞ ∞ ⇐⇒ ∈ ∞ (n)−1 (n) g0 Γ0g0 I s1 = . ⇐⇒ ∩ ∞ 6 ∅

α β Let γ ρ in Γ0 then we have n (n)−1 α β (n) α βT − g0 g0 = n . (A.4) γ ρ ! γT ρ !

On the other hand, for ( a b ) in I we have c d ∞

118 A.2. Representatives for the edges of Γ T 0 \

a b a b 0 1 s1 = c d ! c d ! 1 0 ! b a = . (A.5) d c !

Then from equations (A.4) and (A.5) we have d = γT n. But d O and val(γT n) > ∈ ∞ n > 1 since n> 0, and we get a contradiction. The claim follows.

Remark A.2.3. From Proposition A.2.2 we have

1 a) The ﬁber o1− (JI2K0) has only one element. That is, there is only one edge with origin

JI2K0.

b) If g = I there are exactly two edges with origin JgK , namely, the class JgK and 6 2 0 1 0 1 Jgs1K1 where s1 =( 1 0 ).

The following is a portion of the graph Γ T. 0 \

(0) (1) (2) (3) g0 g0 g0 g0

JΛ0K0 JΛ1K0 JΛ2K0 JΛ3K0 ...

(−1) (−2) (−3) (−4) g0 g0 g0 g0

Corollary A.2.4. The set

n n π− 0 0 π RΓ0 = n > 0 n< 0 (A.6) ( 0 1 ! ) ( 1 0 ! ) [

is a set of representatives for the edges of Γ T. 0 \ R (n) (n) π−n 0 > (n) 0 πn Let us denote the elements of Γ0 by g0 , i.e., g0 = 0 1 for n 0 and g0 =( 1 0 ) I for n < 0. Then given g GL2(K ) there exist n Z, γ Γ0 and κ such that (n) ∈ ∞ ∈ ∈ ∈ ∞ g = γg0 κ.

1 0 Proof. From Proposition A.2.1 we have that the matrix Λn = ( 0 πn ) for n > 0, is a representative for the vertices of X(G1).

(0) 1 If n =0 then Λ0 = g0 is the unique element in o1− (JI2K0).

119 A. Algorithms for the Quotient graph

If n > 1 we have

n (n) 1 0 π− 0 g0 = n n 0 π ! 0 π− ! n π− 0 =Λn n . 0 π− !

Also we have the equality

n ( n) 1 0 0 1 π− 0 − g0 = n n 0 π ! 1 0 ! 0 π− ! n π− 0 =Λns1 n . (A.7) 0 π− !

(n) ( n) Then, by Proposition A.2.2, g0 and g0− are representatives for the two edges with origin

JΛnK0

Remark A.2.5. Given a g in GL2(K ) we can use the algorithm from [But07] to decom- ∞ pose g as g = γΛnα for some n > 0, γ Γ0 and α K . From Proposition A.2.2 we see ∞ n ∈ ∈ that g Jg0 K1 if and only if α I . ∈ ∈ ∞

(n) We explain now how to decompose an element g GL2(K ) as g = γg0 κ for some n Z, ∈ ∞ ∈ γ Γ0 and κ I . ∈ ∈ ∞

Let us consider the class of g in Γ0 GL2(K )/K then \ ∞ ∞

g = γΛmα

1 0 where γ Γ0, α K ,Λm = ( 0 πm ) and m > 0. Write α = rι′ for ι I and r R ∈ ∈ ∞ ∈ ∞ ∈ (m) πm 0 (m) as above (cf. A.2.2) and Λm = g0 ( 0 πm ). Then g = γg0 rι for a unique m > 0 and πm 0 ι = ι′ ( 0 πm ).

1 0 Case 1 If r =( 0 1 ) then we are done.

Case 2 If m = 0, then g(m) =( 1 0 ) and γr Γ and we are done. 0 0 1 ∈ 0

120 A.2. Representatives for the edges of Γ T 0 \

Case 3 If m > 1 we may assume that r =( s 1 ) for some s F . Then we have 1 0 ∈ q π n 0 s 1 π n 0 1 s 0 1 γ − ι = γ − ι 0 1 1 0 0 1 0 1 1 0 π n 0 1 s πn 0 π n 0 0 1 = γ − − ι 0 1 0 1 0 1 0 1 1 0 1 sT n 0 π n = γ − ι 0 1 1 0 n 1 sT ( m) = γ g − ι 0 1 0

The previous discussion is easily transformed into an algorithm which allows us to ﬁnd the

corresponding decomposition of g in the quotient Γ0 GL2(K )/I . \ ∞ ∞

Algorithm 3: decom

Input: A matrix g GL2(K ). ∞ ∈ (n) Output: A list D = [n, γ, κ] such that g = γg0 κ with γ Γ0 and κ I . ∈ ∈ ∞

1: Write g as g = γΛnα for γ Γ0 and α K ⊲ cf.[But07, Lemma 18] ∈ ∈ ∞ 2: if α I then ∈ ∞ 3: define D = [n,γ,α] 4: else 5: if n =0 then c 1 6: search c Fq such that ( 1 0 ) α I ∈ ∈ ∞ 0 1 c 1 7: define D = [n, γ ( 1 c ) , ( 1 0 ) α] − 8: if n =1 then c 1 9: search c Fq such that ( 1 0 ) α I ∈ ∈ ∞ 0 1 c 1 1 10: define D = [ 1,γ ( 1 c ) , δ ( 1 0 ) αδ− ] − − 11: end if 12: if n> 1 then

13: write g as gδ = γΛn 1α − 1 πn−1 0 14: deﬁne D = [ n,γ,δαδ− n−1 ] − 0 π 15: end if 16: end if 17: end if

121 A. Algorithms for the Quotient graph

18: Return D

The function origin (A.2) allows us to deﬁne another function called terminus, as follows

t1 :Γ0 GL2(K )/I Γ0 GL2(K )/K \ ∞ ∞ −→ \ ∞ ∞ JgK t (JgK )= o (JgδK ) 1 7−→ 1 1 1 1 where the matrix δ is the non trivial element in the quotient class of N/I, with N is the normalizer of I in GL2(K ). ∞ We conclude this section by giving some diagrams that are very useful to describe some of the algorithms that are consequence of Algorithm 3. Let us consider ﬁrst

o / GL2(K )/I = Y (T) GL2(K )/K = X(T) ∞ ∞ ∞ ∞

prY prX / Γ0 GL2(K )/I = Y (G1) Γ0 GL2(K )/K = X(G1). \ ∞ ∞ o1 \ ∞ ∞

Where the maps prY and prX are the projections in the quotient graph of edges and vertices, respectively. They are deﬁned as follows

prY : GL2(K )/I Γ0 GL2(K )/I ∞ ∞ −→ \ ∞ ∞ [g] JgK 1 7−→ 1 and

prX : GL2(K )/K Γ0 GL2(K )/K . ∞ ∞ −→ \ ∞ ∞ [g] JgK 0 7−→ 0 It is straightforward to verify that

prX (o([g]1)) = o1(prY ([g]1)). (A.8)

From 2.8 we have that the quotient graph by the congruence subgroup Γ (N) for N § 0 ∈ Fq[T ], is a covering of the tree G1. The functions origin and terminus are also well deﬁned here

oN :Γ0(N) GL2(K )/I Γ0(N) GL2(K )/K \ ∞ ∞ −→ \ ∞ ∞ JgK JgK 1,N 7−→ 0,N

122 A.2. Representatives for the edges of Γ T 0 \

and

tN (JgK1,N )= oN (JgδK1,N ).

We also have well deﬁned projection maps from T to Γ (N) T as follows 0 \

o / GL2(K )/I = Y (T) GL2(K )/K = X(T) ∞ ∞ ∞ ∞

prYN prXN / Γ0(N) GL2(K )/I = Y (Γ0(N) T) o Γ0(N) GL2(K )/K = X(Γ0(N) T), \ ∞ ∞ \ 1,N \ ∞ ∞ \

with the maps prYN and prXN deﬁned as

prY : GL2(K )/I Γ0(N) GL2(K )/I N ∞ ∞ −→ \ ∞ ∞ [g] JgK 1 7−→ 1,N and

prX : GL2(K )/K Γ0(N) GL2(K )/K N ∞ ∞ −→ \ ∞ ∞ [g] JgK . 0 7−→ 0,N

A set of representatives for Γ0(N) GL2(K )/I \ ∞ ∞

We can not deﬁne a canonical set of representatives for the double quotient

Γ0(N) GL2(K )/I . However, after ﬁxing a set SN of representatives for Γ0(N) Γ0, \ ∞ ∞ \ we may use the fact that Γ (N) T is a covering of G to give a non-canonical set of 0 \ 1 representatives for the edges of Γ (N) T. 0 \

Since Γ0 is discrete in GL2(K ), the stabilizers in Γ0 of edges or vertices are finite. More ∞ specifically, define

B0 = GL2(Fq)

and for n > 1

a b B = a, c F×, b F [T ] with deg(f) 6 n . (A.9) n 0 d ∈ q ∈ q

Then for n > 0, Bn is the stabilizer in Γ0 of the vertex represented by Λn in T. Analogously B B is the stabilizer in Γ of the edge with origin [Λ ] and terminal [Λ ] (cf. [Ser03, n∩ n+1 0 n 0 n+1 0

123 A. Algorithms for the Quotient graph

Prop. 3, p. 87]). Note that B B = B for n > 1 and B B is the set of upper n ∩ n+1 n 0 ∩ 1 triangular matrices with entries in Fq.

Let S = s , ..., s be a set of representatives for Γ (N) Γ . Then given γ Γ there N { 1 r} 0 \ 0 ∈ 0 exits a β Γ (N) and a unique s S such that γ = βs . ∈ 0 i ∈ N i

In [But07] there is a method to calculate SN (cf. [Ibid., Lemma 1.22]), actually we also have the following correspondence (cf. [Ibid., Cor. 1.23]).

Proposition A.2.6. Let N F [T ]. Then Γ (N) Γ = P1(F [T ]/N). ∈ q 0 \ 0 ∼ q

Once the set SN is ﬁxed, every g GL2(K ) can be written as ∈ ∞ (n) g = γg0 κ γ Γ0 and κ I ∈ ∈ ∞ (n) = βsig0 κ.

(n) (n) Therefore, given si,sj SN two matrices such that Jsig0 K1,N = Jsjg0 K1,N then there ∈ 1 exists b Bn such that sibsj− Γ0(N). So, for each class in Γ0(N) GL2(K )/I ∈ ∈ \ ∞ ∞ representing an edge e, we choose one representative se of SN and consider the subset S S consisting of a representative for each e G . This set S allows us to deﬁne N ⊂ N ∈ N,M N a set of representatives for Γ0(N) GL2(K )/I as e \ ∞ ∞ e R := s g(n) s S and g(n) R . (A.10) N { e 0 | e ∈ N 0 ∈ Γ0 } e A.3 Lifting cycles to T

The Algorithm 3 may be used to lift a cycle in Γ (N) T to a path in T without backtracking. 0 \ Let C = e˜0, ..., e˜h 1 be a cycle in GN such that t1,N (˜eh 1) = o1,N (˜e0) and t1,N (˜ei) = { − } − o1,N (˜ei+1). Then there exists a sequence of consecutive edges e0, e1..., eh 1 in T such that { − } o(e0) is Γ0(N)-equivalent with t(eh 1) and prY (ei)=˜ei for all i in 0, 1, ..., h 1 . − N { − } It is enough to see how to lift two consecutive edges in the cycle C to two consecutive edges

on the tree T. Let us suppose that gi GL2(K ) with JgiK1,N =e ˜i and ei T is a lifting ∈ ∞ ∈ ofe ˜i. We want to ﬁnd a matrix gi+1 GL2(K ) such that it liftse ˜i+1 to an edge ei+1 T ∈ ∞ ∈ with t(ei)= o(ei+1).

Claim: The matrix g = g δτ δ for some a F were δ =( 0 1 ) and τ =( π a ). i+1 i a ∈ q π 0 a 0 1

124 A.3. Lifting cycles to T

We have that t([gi]1)= o([giδ]1) and multiplication by τa, gives on the tree T all the edges

whose terminal is the vertex o([giδ]1) (see ﬁgure, the blue edges corresponds to the wanted path) τa

[gi+1]1 [gi]1 q edges [giδ]1

[giδτa]1

that is o([giδ]1)= t([giδτa]1), by deﬁnition of the function origin and terminal we have

t([gi]1)= o([giδτaδ]1).

Hence the q edges with origin t(e ) are given by [g δτ δ] for all a F . So one of these i i a 1 ∈ q edges project toe ˜i+1 or equivalently prYN ([giδτaδ]1)=˜ei+1, which proves the claim.

We can summarize the previous discussion with the following algorithm.

Algorithm 4: FindPath

Input: A list C = [g0,g1, ..., gr 1] of length r of matrices in GL2(K ) representing a − ∞ cycle in Γ (N) T, that is, t (Jg K ) = o (Jg K ) for all i 0, ...r 2 and 0 \ N i 1,N N i+1 1,N ∈ { − } tN (Jgr 1K1,N )= oN (Jg0K1,N ). − Output: A list P = [h0, h1, ..., hr 1] of length r of matrices in GL2(K ) representing a − ∞ path in T and such that o([h0]1) is Γ0(N)-equivalent to t([hr 1]1). −

1: Make a list L with the elements of the ﬁnite ﬁeld Fq

2: set P = [g0] 3: for i from 1 to r 1 do − 4: let n = #P and set gaux = P[n] ⊲ gaux is the last element of P 5: bool false ← 6: k 1 ← 7: while bool=false do 8: a L[k] ← 9: set haux = gauxδτaδ

125 A. Algorithms for the Quotient graph

10: if pr1,N ([haux]1)= gi+1 then 11: bool true ← 12: end if 13: k k +1 ← 14: end while

15: append haux to P 16: end for 17: Return P

A.4 Finding the representative

In many algorithms, it is necessary to work with elements in GL2(K ) which belong to the ∞ same class in Γ0(N) GL2(K )/I . So given two different matrices g1 and g2 in GL2(K ) \ ∞ ∞ ∞ such that Jg1K1,N = Jg2K1,N , we want to find γ Γ0(N) and κ I such that g1 = γg2κ. ∈ ∈ ∞ Using Algorithm 3 we can give an efficient method to write such decomposition. We present first the following result as a trivial corollary of Proposition A.2.4. Nevertheless we give a constructive proof which gives Algorithm FindTheRep.

Corollary A.4.1. Let g1,g2 GL2(K ). Suppose that Jg1K1,N = Jg2K1,N . Then there is ∈ ∞ an algorithm to ﬁnd γ Γ0(N) and κ I such that g1 = γg2κ. ∈ ∈ ∞

Proof. Let g1,g2 GL2(K ) be two matrices Γ0(N)-equivalent. Then g1 and g2 are in the ∈ ∞ same class in GL2(A) GL2(K )/I , therefore there exist s1,s2 GL2(A) and κ1, κ2 I \ ∞ ∞ ∈ ∈ ∞ such that

(n) g1 = s1g0 κ1, (n) g2 = s2g0 κ2

(n) where g0 is the representative for the quotient GL2(A) GL(K )/I described in Propo- \ ∞ ∞ sition A.2.4.

126 A.4. Finding the representative

Since g1 and g2 are Γ0(N)-equivalent we have

(n) (n) (n) (n) Γ0(N)s1g0 I =Γ0(N)s2g0 I s1g0 Γ0(N)s2g0 I ∞ ∞ ⇐⇒ ∈ ∞ 1 (n) (n) 1 I2 s1− Γ0(N)s2g0 I (g0 )− ⇐⇒ ∈ ∞ 1 (n) (n) 1 s1− Γ0(N)s2 g0 I (g0 )− = ⇐⇒ ∩ ∞ 6 ∅ 1 (n) (n) 1 s1− Γ0(N)s2 Γ0 g0 I (g0 )− = . ⇐⇒ ∩ ∩ ∞ 6 ∅ An easy calculation shows that

(n) (n) 1 Γ0 g0 I (g0 )− = Bn, ∩ ∞ where Bn is the set deﬁned in (A.9). Then there exists a b Bn and γ Γ0(N) such that 1 ∈ ∈ s1− γs2 = b. A direct calculation shows that setting

1 (n) 1 1 (n) κ = κ2− (g0 )− b− g0 κ1

1 satisfy g1 = γg2κ. We only need to verify that κ is an element of I . Indeed κ2− and κ1 ∞ (n) (n) 1 (n) 1 1 (n) are in I , and since b g0 I (g0 )− we have (g0 )− b− g0 I . ∞ ∈ ∞ ∈ ∞

Algorithm 5: FindTheRep

Input: Two matrices g1 and g2 in GL2(K ) that are Γ0(N)-equivalent. ∞ Output: Two Matrices γ Γ0(N) and κ I such that g1 = γg2κ. ∈ ∈ ∞

1: Use Algorithm 3 to write

(n) g1 = s1g0 κ1 (n) g2 = s2g0 κ2

2: calculate the stabilizer Bn 3: bool false ← 4: i 1 ← 5: while bool=false do 6: b B [i] ← n

127 A. Algorithms for the Quotient graph

1 7: if s bs− Γ (N) then 1 2 ∈ 0 8: bool true ← 9: end if 10: i i +1 ← 11: end while 12: Return

1 γ = s1bs2− and 1 (n) 1 1 (n) κ = κ2− (g0 )− b− g0 κ1.

128 B. Algorithms for the table

In this appendix we describe the main algorithms that we use in the calculation of the table and the integral. The ﬁrst section deals with the implementation of the Hecke operator U , ∞ the algorithm that allows us to build up the table and the one that we use to decompose

functions in FI as a product of elements in the pseudo-basis BM .

In the second section we give the algorithm to calculate the valuation of the integral and the one to lift a path in the tree T to a path in Ω. Finally we explain how to calculate the Tate parameter and give a proof of Theorem 5.6.1.

B.1 Algorithms for the calculation of the table

The Hecke Operator U ∞

We recall the deﬁnition of the Hecke operator

(U φ)(γ)(f(t)) := φ (γτa) f(πt + a) ∞ a Fq Y∈

for φ S, f FI and γ Γ P GL2(K ). ∈ ∈ ∈ \ ∞ A crucial fact that helps us to calculate U quickly is the I -equivariance. As we already ∞ ∞ explained in Chapter 4, it is enough to calculate the values of U in a set of representatives ∞ RN of the edges of the quotient Γ0(N) GL2(K )/I up to some level M. \ ∞ ∞ Therefore for each γ R rep ∈ N δ i j δ i j (U φ)(γrep)(1 + ξ π t ) := φ (γrepτa)(1+ ξ π (πt + a) ). (B.1) ∞ a Fq Y∈

129 B. Algorithms for the table

From the deﬁnition of the operator U , we have that for each a Fq, we need to ﬁnd the ∞ ∈ edge of GN,M corresponding to γrepτa. However, we only know the values of φ in elements of

RN . Via Algorithm 5, we ﬁnd the corresponding representative of γrepτa in RN as follows: we consider γrepτa as an element of GL2(K )/I and use the projection function prY to ∞ ∞ N ﬁnd ga GL2(K ) such that ∈ ∞

JgaK1 = JγrepτaK1

= JγˆrepK1 for someγ ˆrep RN . Then using Algorithm 5, we can ﬁnd γ Γ0(N) and κ I such that ∈ ∈ ∈ ∞

γrepτa = γγˆrepκ.

Plugging in this in equation (B.4) we get

δ i j δ i j (U φ)(γrep)(1 + ξ π t )= φ (γrepτa)(1+ ξ π (πt + a) ) ∞ a Fq Y∈ δ i j = φ (γγˆrepκ)(1+ ξ π (πt + a) )

a Fq Y∈ = φ (ˆγ )(κ (1 + ξδπi(πt + a)j)). rep ∗ a Fq Y∈

In the last equality we use the Γ0(N)-invariance of the function φ, Lemma 4.3.7 and the action deﬁned in (4.6). From the preceding discussion we see that for a ﬁxed γ R in rep ∈ N order to evaluate φ(γ )(f ) for f B we need to calculate γ τ in many instances. rep ij ij ∈ M rep a

Remark B.1.1. We save time by precalculating the decomposition γrepτa = γγˆrepκ for all a F and γrep R . So for each γrep we store in the data structure of the quotient graph ∈ q ∈ N GN,M the list [γrep, a, γˆrep, κ] (called “signature”) to indicate that γrepτa = γγˆrepκ for all a F . ∈ q

Therefore the computation of the operator U breaks up in two parts. The ﬁrst one is ∞ carried out with the quotient graph GN,M and consists in the elaboration of the list with the signatures [γrep, a, γˆrep, κ] and the second one takes place during the calculation of the table and reduces to read the signatures from a list.

Observe that the calculation of the operator U does not depend on the harmonic cocycle ∞ explicitly.

130 B.1. Algorithms for the calculation of the table

The table

Let ϕ Hnew(T, Z)Γ0(N), as we have seen in 3.1.1, the harmonic cocycle ϕ gives rise ∈ ! § 1 Γ to a measure µϕ on M0(P (O ), Z) . The objective of this subsection is explain how to ∞ calculate the table that allows us to evaluate any integral of the form

f dµ × ϕ Z O∞ for f F with accuracy up to πM , where M is an integer > 1. ∈ I In 4.3 we explained already how to ﬁll out this table, however, the construction given § there is slightly diﬀerent from the actual implementation. Since this is the main algorithm of the whole work, we think it is worthy to explain the actual implementation.

Let us recall some facts and deﬁnitions from 4.3. Let M > 1 be an integer, the associated § pseudo-basis is (cf. Lemma 4.3.1)

B := 1+ ξδπitj (i, j) F , i M, δ =0, ..., d 1 M { | ∈ I ≤ − }

with d the degree of the extension Fq2 over Fp.

A straightforward calculation shows that the cardinality of BM is

(M + 1)(M + 2) d d 1 = (M 2 +3M). 2 − 2

For each γ R we want to calculate rep ∈ N

δ i j δ i j 1 Φ (γ )(1 + ξ π t )= (1 + ξ π t ) d(γ− µ ). (B.2) µϕ rep × rep ∗ ϕ Z O∞ As we discussed in 4.3, for each γ R and a ﬁxed δ we have to calculate 1 (M 2 +3M) § rep ∈ N 0 2 integrals, hence to each γrep we attach d “triangular matrices” Tγrep,δ for varying δ) whose entries are the integrals of the form (B.2) (for functions f of BM,δ (cf. (4.10)). For the implementation we condense all these matrices in one matrix T of size #R #B . N × M

First order the pseudo-basis BM increasingly, according to the order relation deﬁned in 4.3. Observe that the constants correspond to the functions of B with j = 0. There are § M exactly Md constants and their integrals are located in the ﬁrst Md columns of the matrix

131 B. Algorithms for the table

T. The algorithm consists of two stages. In the ﬁrst stage we calculate the values of the function Φµϕ (γrep) at the constants, this can be made very quickly using the formula

δ i δ i ϕ(γrepe0) Φµϕ (γrep)(1 + ξ π )= 1+ ξ π . (B.3) In the second stage, we calculate the value of the integral at the non-constant functions, starting from the smallest one in BM . Note that we start with functions of the form 1+ ξδπM tj for δ 0, ..., d 1 an j from 0 to M. In each case the resulting functions ∈ { − } after applying the U operator factorize as a product of constants, and the value of the ∞ integral is already known. After this is done for all representatives in RN , we proceed to δ M 1 j apply U to functions of the form 1+ ξ π − t , then the exponents of π increase in 1. So ∞ the new factors appearing are either constants or of the form 1 + ξδπM tj. In each case we know the integral.

Continuing in this fashion, we apply the U operator to each element of BM as many times ∞ as necessary, until we get the integral to the wanted precision (cf. 4.3). § δ i j Let us suppose that we want to calculate U (φ)(γrep)(1 + ξ π t ) and suppose that γrep ∞ represents an edge of level l over a cusp of GN,M . Then by deﬁnition of the U we have ∞

δ i j δ i j (U φ)(γrep)(1 + ξ π t ) := φ (γrepτa)(1+ ξ π (πt + a) ). (B.4) ∞ a Fq Y∈

So after applying the operator U we need to calculate over the edges given by γrepτa of ∞ level l 1 at a function f with val (t)= i + 1. Observe that each time that we apply the − π operator U we move over the cusp in direction the compact part of GN,M . The integral is ∞ 1 if after applying successively the operator U we are still over the cusp and the valuation ∞ in π of the function is M, this happens if and only if

i + l > deg(N)+ M.

Remark B.1.2. We use the previous inequality to rule out in the calculation of the table the functions and edges for which we know a priori that the integral is 1.

132 B.1. Algorithms for the calculation of the table

Algorithm 6: TheTable Input: The data structure G corresponding to the quotient graph Γ (N) T up to level N,M 0 \ M. A harmonic cocycle ϕ with rational Hecke eigenvalues. This is actually a list with

all the values of ϕ at the edges of GN,M .

Output: A matrix T with entries in 1 + πFq2 JπK with all the values of the integrals of M functions in BM . (The entries of T are elements in 1 + πFq2 JπK modulo π ).

1: Set R to be the list of representatives of the edges of the graph Γ (N) T up to depth N 0 \ M 2: initialize a matrix T of ones of size n m, where n = #R and m = #B × N M 3: for s from 1 to n do 4: for k from 1 to Md do ⊲ the constants are the last Md entries 5: T[s][k] B [k]ϕ(RN [s]e0) ⊲ Here we ﬁll the entries corresponding to the ← M constants. 6: end for 7: end for 8: for s from 1 to n do 9: for k from Md +1 to m do 10: f B [k] ← M 11: set γ R [s] rep ← N 12: set level(γrep) the level of the edge given by γrep ⊲ apply Algorithm 3 13: set ii = val (f 1) π − 14: if ii + level(γ ) deg(N) < M then ⊲ cf. Remark B.1.2 rep − 15: set I 1 ← 16: for a Fq do ⊲ we start to apply the U operator ∈ ∞ 17: read from GN,M the signature [γrep, a, γˆrep, κ] 18: s Position (R , γˆ ) 0 ← N rep 1 19: f f τ κ− ← h a i 20: set a list Lf with the factors of f ⊲ apply Algorithm 7

21: for l in Lf do 22: k Position (B ,l[1]) 0 ← M 23: I I T[s ][k ]l[2] ⊲ the value T[s ][k ] is known ← × 0 0 0 0 24: end for

133 B. Algorithms for the table

25: end for 26: T[s][k] I ← 27: end if 28: end for 29: end for 30: Return T

Remark B.1.3. 1. Note that after ﬁlling the columns that correspond to the constants, we continue ﬁlling the table T from the “left to the right”. In the step 16 we apply the Hecke operator U . So if we are in the column j-th we only need values of the ∞ integral already stored in other columns on the left.

2. We do not need to give in the input the integer M since the basis BM is stored in the

data structure of GN,M .

new Γ0(N) Let us suppose that there are h elements in H! (T, Z) with rational Hecke eigenvalues. In this case there are h isogeny classes of the given conductor N . To ﬁnd these classes ∞ we need to calculate h integrals and therefore h tables. We save time if we make all the tables simultaneously as follows.

Let us assume that H (T, Z)Γ0(N) has dimension g > h. Let B := ψ , ..., ψ be its ! { 1 g} standard basis as deﬁned in 2.9. Let also ϕ , ..., ϕ be the set of harmonic cocycles with § { 1 h} rational Hecke eigenvalues, then

g ϕ = c ψ for c Z. ij j ij ∈ j=1 X Since the integral is multiplicative, we have

z0 t cϕi (γ)= − dµϕi (t) ×∂Ω γz0 t Zg − z t cj = 0 − dµ (t) . × γz t ψj j=1 ∂Ω 0 Y Z − So we can make tables for all the ψ’s instead of calculate many tables for each harmonic cocycle with rational Hecke eigenvalues. The reason is that for each entry of the table

134 B.1. Algorithms for the calculation of the table

we need to calculate the operator U and it does not depend on the harmonic cocycle ∞ explicitly.

For the implementation we take Algorithm 6 we need to initialize g matrices of 1’s instead of only one and modify the steps 3-7 to calculate the values for the integral at the constant at all the elements in the standard basis instead of doing it only for ϕ.

Analysis for the algorithm TheTable

For the cost analysis of the algorithm to calculate the table, we need to consider separately two loops. The ﬁrst one is the loop to calculate the integral at the constants and the second one is the part in which we apply the operator U . ∞ Before starting with the analysis for the running time of Algorithm 6, we need a bound for the number of edges of the quotient graph GN,M . We know that #GN,M = #CN + #SPN , where CN and SPN are the compact part and the cusps of GN,M , respectively.

Unfortunately there is no easy formula for the compact part. However we can use the fact that the quotient graph GN,M is a covering of G1 to estimate the number of edges in the compact part as #S deg(N), where S is a set of representatives for Γ (N) Γ . From N N 0 \ 0 s ri [Non01, p. 68] we have that if N = i=1 fli with fli a prime polynomial of degree li then Q s li(ri 1) li #SN = q − (q + 1) i=1 Y and κ(N) 2s #SP =2s + − N q 1 −

s li (ri 1)/2) li ri/2) where κ(N) := i=1 q ⌊ − ⌋ + q ⌊ ⌋ . Then #SN is a polynomial in q of degree C s deg(N) SP deg(N)/2 s deg(N). A boundQ for # N is 2 q deg(N ) and for # N is q 2 . Therefore the number of edges in G(N, M) is bounded by

2sqdeg(N) deg(N)+2sMqdeg(N)/2. (B.5)

For the ﬁrst loop (steps 3-7), note that we are exponentiating polynomials in π of the form 1 + ξsπi for 1 6 i 6 M and this can be done via binomial expansion (1 + ξsπi)k = k k s l il M+1 l=0 l (ξ ) π and using the fact that we are working modulo π , so that is suﬃces to P

135 B. Algorithms for the table

M compute for l in the range 0, ..., i is bounded by M/i. Summing over the two loops the number of operations is bounded by

d 1 M − n M/i 6 ndM(1 + log M) i=1 Xδ=0 X M using that i=1 1/i 6 1 + log M and where n = #GN,M . Hence the cost for all iterations of the loopP is 2sMd (1 + log M) qdeg(N) deg(N)+2sMqdeg(N)/2 . (B.6) Examining the second loop shows that the number of operations inside it, depends on the condition “if” (step 14) and the “for” loop in step 16. So we need to count the number of edges in G and functions in B that satisfy the inequality i+level(γ ) deg(N) < M. N,M M rep − Let us ﬁrst to consider some particular cases. If i = M then the inequality becomes level(γrep) < deg(N), then there are #CN Md edges and functions that verify the condition in step 14.

If i = M 1 then we have level(γ ) 1 < deg(N), in this case there are (#C +#SP )(M − rep − N N − 1)d edges and functions that satisfy the inequality. We can see here that for i = M k − the inequality becomes level(γ ) k< deg(N) then there are (#C + k#SP )(M k)d. rep − N N − Therefore we can have that the number of edges and functions that verify the condition in step 14 is

#C Md + ... + (#C + k#SP )(M k)d + ... + (#C +(M 1)#SP )d. N N N − N − N

A straightforward calculation shows that the last expression simpliﬁes to

#C #SP d N (M 2 + M)+ N (M 3 M) . (B.7) 2 6 − The loop that starts in step 16 is executed q times and it includes the function to factorize which has a cost of 5M 4 (see the analysis for Algorithm 7). Hence the cost for all iterations of the second loop is 5dqM 4 #CN (M 2 + M)+ #SPN (M 3 M) . Which is also a bound 2 6 − for the cost of the algorithm, since the total of operations in the algorithm is the sum of the costs in the two loops but the dominant cost of the algorithm is the second loop. We may summarize the previous discussion in the following proposition.

136 B.1. Algorithms for the calculation of the table

Proposition B.1.4. The Algorithm 6 can be performed using no more that

5 7 s n +1 5 s 6 n+1 1 n+1 s 5 s n +1 5 dM 2 q 2 + 2 dM q n + d 5q 2 n 2 q 2 M 6 2 2 − 3 operations, where n is the degree of N.

Factorization

We now consider the problem of factorizing functions in FI . In Chapter 4 we stated the following lemma, here we give a constructive straightforward proof that will lead us to the algorithm of factorization.

Lemma B.1.5. Given any function f F and any integer M > 1, then there exists a ∈ I ﬁnite set of indices J I and m 1, 2, ..., p 1 such that ⊆ ijδ ∈{ − } f (1 + ξδπitj)mijδ (mod πM+1) ≡ (i,j) J Y∈ where ξ F 2 is a primitive element for the extension over F , δ 0, ..., d 1 with d ∈ q p ∈ { − } the degree of the extension Fq2 over Fp. The representation is unique modulo p powers of δ i j fijδ =1+ ξ π t , so the set

B := 1+ ξδπitj (i, j) F , i M, δ =0, ..., d 1 . M { | ∈ I ≤ − } is what we call a multiplicative pseudo-basis.

i j Proof. For f = (i,j) I aijπ t Fq2 JtKJπK we set ∈ ∈ P val (f) = min i > 0 j > 0 such that a =0 π { | ∃ ij 6 }

i j (with min∅ = ) and for i > 0 we deﬁne f[π ] := a t , furthermore for g F 2 JtK, ∞ j ij ∈ q val denotes the usual valuation in t. t P Let f =1+ a πitj in F , let i = val (f 1) and j = val (f[πi0 ]). Then we can (i,j) ij I 0 π − 0 t write f as P i0 j0 i j f =1+ ai0j0 π t + aijπ t .

(i,j)=(i0,j0) I 6 X ∈

137 B. Algorithms for the table

2 d 1 d 1 s 2 − Note that since 1,ξ,ξ , ..., ξ − is a basis of Fq over Fp we have ai0j0 = s=0 bsξ for b F . Using this, a straightforward calculation shows that s ∈ p P

d 1 d 1 bs − − bs (1 + ξsπi0 tj0 )bs = (ξs)k(πi0 tj0 )k (B.8) k s=0 s=0 Y Y Xk=0 d 1 − s i0 j0 i0+1 = (1 + ξ bsπ t + O(π )) s=0 Y d 1 − s i0 j0 i0+1 =1+ bsξ π t + O(π ) s=0 ! X 1+ a πi0 tj0 (mod πi0+1). ≡ i0j0

d 1 s i j b M+1 Set f = − (1 + ξ π 0 t 0 ) s F . Dividing f by f and reducing modulo π we i0j0 s=0 ∈ I i0j0 get Q

f0 = f/fi0j0 =(f +(f f )/f ) i0j0 − i0j0 i0j0 1 =1+(f f )f − − i0j0 i0j0 i0 j i j =1+ bi0jπ t + aijπ t

j>j0 i>i0 X jX>j0

I.e., val (f 1) > i and val (f [πi0 ]) > j . Note that the factors 1+ ξsπi0 tj0 of f are π 0 − 0 t 0 0 i0j0 elements of the pseudo-basis BM .

Continuing in this fashion, since the function f F , we can construct a sequence f F 0 ∈ I k ∈ I valπ(f 1) with (1 + M)val (f 1) + val (f [π k− ]) strictly increasing and then the procedure π k − t k finishes after a finitely number of divisions since we are working modulo πM+1. Defining J to be the set consisting of indices (i , j ) such that i = val (f 1) and j = val (f [πik ]) k k k π k − k t k coming from the step k, we have that

f (1 + ξδπitj)mijδ (mod πM+1) ≡ (i,j) J δ 0,...,d 1 Y∈ ∈{ Y − } which follows by construction and equality (B.8).

Remark B.1.6. Note that the factorization of a function f BM is not unique. Let δ i j 2 ∈ fij = 1+ ξ π t BM be a factor of a function f with q gcd(i, j) then it admit other ∈ |2 representation as a product of elements of BM , namely the q -th roots of fij.

138 B.1. Algorithms for the calculation of the table

We conclude with Algorithm 7 which is basically a transcription of the proof of Lemma B.1.5.

Algorithm 7: factorize

Input: A function f in FI and an integer M. Output: A list L consisting of pairs [f , e ] with f B , e > 0 and (i, j) J I ij ij ij ∈ M ij ∈ ⊂ fij M+1 such that f f L fij (mod π ). ≡ ij ∈ Q 1: Initialize L = [ ] M+1 2: define faux = f (mod π ) 3: while f =1 do aux 6 4: i val (f 1) 0 ← π aux − 5: j val (coefficient(i , f 1)) 0 ← t 0 aux − 6: write f =1+ a πi0 tj0 + a πitj aux i0j0 (i,j)=(i0,j0) I ij 6 ∈ d 1 s 7: write a = − b ξ with b F i0j0 s=0 s Ps ∈ p i0 j0 8: set z = π t P ⊲ we change the variable since i0 and j0 are fixed d 1 s b − s 9: set fi0j0 = s=0(1 + ξ z) ⊲ use the binomial expansion 1 i j 10: compute g = f − ⊲ We make the change of variables again z by π 0 t 0 Q i0j0 11: f f g aux ← aux s i0 j0 12: append to L all the pairs [1 + ξ π t , bs] from the factorization of fi0j0 (step 9) 13: end while 14: Return L

Analysis for the algorithm factorize

A closer look to the algorithm shows that in order to ﬁnd the running time, it is enough to ﬁnd a bound for the number of iterations of the while-loop (step 3-step 13) in the worst case, which occurs when the function f is divisible by all the elements of the pseudo-basis B B d 2 M . A straightforward calculation shows that # M = 2 (M +3M). Step 7 can be done in O(d) and its cost is negligible since d is constant and M is larger d 1 s i j b − 0 0 s 6 than d. In step 9 we calculate the product fi0j0 = s=0(1 + ξ π t ) with 1 bs < p, Q

139 B. Algorithms for the table

therefore since πi0 tj0 is ﬁxed, we can make the substitution πi0 tj0 by z, before to carry out the exponentiation, then we get (1 + ξsz)bs so we can use the binomial expansion to obtain bs bs s l l l=0 l (ξ ) z . Since bs < p we have that this exponentiation can be performed in at s l d 1 mostP p operations, we need to express (ξ ) in the basis 1, ..., ξ − over Fp. On the other hand, after exponentiating we get d polynomials of degree at most p in the new variable z and multiplying them cost at most (d 1)pM because any partial product has at most M − terms. Hence the cost for the step 9 is pdM.

In step 10 we invert f , again using the auxiliary variable z = πi0 tj0 . Let e = M/i i0j0 ⌊ 0⌋ then

d 1 − 1 δ bs − − g = fi0j0 = (1 + ξ z) δ=0 Yd 1 e − b = (ξδ)l − s zl l Yδ=0 Xl=0 d 1 e − b + l 1 ( ξδ)l s − zl (mod ze). ≡ − l ! Yδ=0 Xl=0 The cost to set up each factor is e, we then carry out d 1 multiplications modulo ze, i.e., − the cost here is (d 1)(e + 1)2. In total the cost is atmost d(e2 +3e + 1). − 2 To compute fauxg, observe that faux has at most M terms and g has e + 1 terms. So the cost for the steps 7-13 is at most

dpM + d(e2 +3e +1)+(e + 1)M 2. (B.9)

We need to sum this over all (i, j) I to obtain the bound C for the number of operations. ∈ M i M 2 3M M C = dpM + d + +1 d + +1 M 2 i2 i i i=1 j=0 ! X X M i M 2d 3Md M 3 = (dpM + d + M 2)+ + + i2 i i i=1 j=0 ! X X M M 2d 3Md M 3 = (dpM + d + M 2)(i +1)+ + + (i + 1) i2 i i i=1 X M 2 +3M M M 2d 3Md M 3 =(dpM + d + M 2) + + + (i + 1) 2 i2 i i i=1 X

140 B.2. Algorithms for the calculation of the integral

M 1 We know from calculus that the harmonic series diverges and i=1 i < 1 + log N. Also 1 π2 M 1 the series ∞ = , then < 2. Let us suppose also that dp < M and M > 6 i=1 i2 6 i=1 i2 P then 4 + logPM < M. ThereforeP we have

3 1 3 3 13 C < M 4 + dp M 3 + M 2 + M 3 + dM 2 + (1 + log M)(dM 2 +3dM + M 3) 2 2 2 2 2 3 1 3 3 13 < M 4 + M M 3 + M 2 + M 3 + dM 2 + (1 + log M)(dM 2 +3dM + M 3) 2 2 2 2 2 5 15 =2M 4 + (4 + log M)(M 3 + dM 2 +3dM)+ dM 2 dM 2 − 2 5 15 < 2M 4 + M(M 3 + dM 2 +3dM)+ dM 2 dM 2 − 2 11 15 =3M 4 + d M 3 + dM 2 M 2 − 2 11 15 < 4M 4 + M 3 M 2 (since d < M) 2 − 2 < 5M 4

Proposition B.1.7. Suppose that M > 6 and dp

B.2 Algorithms for the calculation of the integral

Valuation of the integral

To obtain the Tate parameter associated to an harmonic cocycle ϕ we need to calculate the integral

z t c (γ)= 0 − dµ (t) ϕ × γz t ϕ Z∂Ω 0 − where γ is an element of Γ (N) associated to a cycle c of the quotient graph Γ (N) T 0 0 \ (see 4.4.3). As we mentioned in 5.6 if C = c , ...c is a basis for the homology of the § § { 1 g} quotient graph Γ T then to each c C there is associated a matrix γ Γ, let us call by \ i ∈ i ∈ abuse of notation C the set of all γi’s. Then

val(q) = min val(c (γ )) γ C . { ϕ i | i ∈ }

141 B. Algorithms for the table

Let us consider a cycle c in the quotient tree and v0, v1, ..., vr be a path in T that lifts c.

This sequence of the vertices induces a sequence z0, ..., zr in Ω such that λ(zi)= vi, then

z0 t cϕ(γ)= − dµϕ(t) (B.10) ×∂Ω γz0 t Z r 1 − − z t = i − dµ (t) × z t ϕ Z∂Ω i=0 i+1 − r 1 Y − z t = i − dµ (t). × z t ϕ i=0 ∂Ω i+1 Y Z −

We can use this decomposition to know the valuation of the integral, namely,

r 1 z t − z t val 0 − dµ (t) = val i − dµ (t) × γz t ϕ × z t ϕ Z∂Ω 0 − Z∂Ω i=0 i+1 − ! r 1 Y − z t = val i − dµ (t) × z t ϕ i=0 Z∂Ω i+1 − ! r 1 Y − z t = val i − dµ (t) . (B.11) × z t ϕ i=0 ∂Ω i+1 X Z −

Therefore we need to calculate the valuation of z t i − dµ (t). × z t ϕ Z∂Ω i+1 − We have

µϕ(U) zi t zi t val − dµϕ(t) = val lim − ×∂Ω zi+1 t α zi+1 t −→ U Cα ! Z − Y∈ − z t µϕ(U) = lim val i − α zi+1 t −→ U Cα ! Y∈ − zi t = lim µϕ(U) val − (B.12) α zi+1 t −→ U Cα X∈ − = lim µϕ(U) (B.13)

−→α U Cα X∈ = µ(Uei )

= ϕ(ei)

142 B.2. Algorithms for the calculation of the integral

where ei is the edge with origin vi. The equality from (B.12) to (B.13) follows from the fact that zi and zi+1 are the liftings of consecutive vertices and the point t belongs to one zi t of the open sets in the partition determined by ei, which implies val z − t = 1. Using i+1− the last equality and the equation (B.11) we have that z t val 0 − dµ (t) = ϕ(e ) × γz t ϕ i ∂Ω 0 e Z − Xi

where ei runs over the cycle c.

Summarizing we have that the valuation of the Tate parameter is given by

val(q) = min ϕ(e) c C . | i ∈ (e c ) X∈ i So to calculate the Tate parameter, we choose the γ that lifts the cycle with the minimal valuation. One can obtain this result applying the equation (1.3) from [GEK97].

Lifting vertices to Ω

Let c be the cycle that gives the minimal valuation of the integral for a given harmonic cocycle ϕ and let P = w ,w , ..., w be the lifting of the vertices of c to the tree T to { 0 1 r} consecutive vertices and such that v0 and vr are Γ0(N)-equivalent. That is, there exists

γ Γ0(N) and α K such that w0 = γwrα (they represent vertices not edges). ∈ ∈ ∞ From the discussion above, we need an algorithm to ﬁnd the sequence z , ..., z Ω. This 0 r ∈ can be done using the GL2(K )-equivariance of the reduction map λ : Ω T (cf. 2.6) ∞ −→ § as follows.

Let v0 be standard vertex in T, we know that the standard aﬃnoid is deﬁned to be

1 λ− (v0)= z C× z 1, z c 1 c Fq . ∈ ∞ || | ≤ | − | ≥ ∀ ∈ 1 Let ξ F 2 F then ξ λ− (v ). ∈ q \ q ∈ 0 1 Since the translates of the affinoid λ− (v0) by GL2(K ) cover Ω and two translates are ∞ either identical, disjoint or intersect each other, we can get any affinoid by translating the 1 standard one. Equivalently, given a vertex v T with v = v , to get a element in λ− (v) it ∈ 6 0 1 is enough to find a g GL2(K ) such that [v]0 = [gv0]0. Then we have that g ξ λ− (v). ∈ ∞ h i ∈

143 B. Algorithms for the table

There is no loss of generality in assuming that the matrices wi are in normal form. Let say πk u > that w0 = 0 1 for some k 0. Then we have that

k 1 z = π ξ + u λ− (w ). (B.14) 0 ∈ 0

Since we can go from wi to the next vertex wi+1 by multiplying by an appropriate element gi GL2(K ), we may deﬁne zi+1 = gi zi , i.e., the action of gi in zi by M¨obious ∈ ∞ h i transformations.

On the other hand, to calculate our integral we use the equality zr = γz0, however, on the

tree we have [vr]0 = [γv0]0, so the elements gi are required so, that its product is γ. The

following algorithm allows us to ﬁnd the sequence of the gi’s with such property.

Algorithm 8: TransitionGammas

Input: A list P = [w0,w1, ..., wr] of matrices in GL2(K ) representing consecutive vertices ∞ in T and such that w0 and wr are Γ0(N)-equivalent.

Output: A list L consisting of matrices gi GL2(K ) such that [giwi]0 = [wi+1]0 and ∈ ∞ the product of the gi’s is in Γ0(N).

1: Write wr = γw0α with γ Γ0(N) and α K . (Use Algorithm 4) ∈ ∈ ∞ 2: make a loop to deﬁne the list

1 1 1 1 1 L = [w1wr− γ,w2w1− ,w3w2− , ..., wr 1wr− 2,wrwr− 1] − − −

3: Return L

A direct calculation shows that the product of the elements in L is actually γ. Also the condition

[giwi]0 = [wi+1]0 is veriﬁed. Using Algorithm 8 we can now lift the vertices of a given path to a sequence of zi’s in Ω.

Algorithm 9: LiftToOmega

144 B.2. Algorithms for the calculation of the integral

Input: A list P = [w0,w1, ..., wr] of matrices in GL2(K ) representing consecutive vertices ∞ and such that w0 and wr are Γ0(N)-equivalent. Output: A list Z consisting of diﬀerent elements in Ω lifting the vertices of P.

1: Use Algorithm 8 to produce the list S = [g0,g1, ..., gr 1] − 2: initialize Z = [z = w ξ ] ⊲ We lift w as in (B.14) 0 0h i 0 3: for i from 2 to r do

4: append to Z the element gi 1 Z[i 1] ⊲ is the action by M¨obius − h − i h·i transformation. 5: end for 6: Return Z

The integral

In Chapter 4 we saw how to carry out the change of variables, and we explained how we can

integrate over an edge e determined by the adjacent vertices v and v′. Using the partition induced by Lemma 4.4.1, we can break up the integral t z − dµ (t) × t z ϕ Z∂Ω − ′ where z and z′ are liftings to Ω of v and v′, respectively. For the change of variables we k also supposed that v = [k,u] and v′ = [k +1,u+a π ] for some a F , respectively. From 0 0 ∈ q k Lemma 2.4.5 we have that all the neighbors of v different from v′ are given by [k+1,u+aπ ] k 1 for a Fq a0 and [k,u mod π − O ], which is the only one that can not be obtained ∈ \ ∞ k k+1 form v multiplying by τa. The neighbors of v′ are of the form [k +2,u + a0π + bπ ] with b F . ∈ q k 1 Notation: All the neighbors of the vertices v and v′ different from [k,u mod π − O ] ∞ k 1 are called the neighbors of zero, while [k,u mod π − O ] is called the neighbor of infinity. ∞

Algorithm 10: IntegrateOverEdge k Input: An edge e in GN,M given by the vertices v = [k,u] and v′ = [k +1,u + a0π ]. A

pair [z, z′] of elements in Ω over o(e) and t(e), respectively. A harmonic cocycle ϕ. The

145 B. Algorithms for the table

data structure G of the quotient graph Γ (N) T. The table T with the values of N,M 0 \ the integral at the functions of BM . Output: The integral t z − dµ (t) × t z ϕ Z∂Ω − ′ over the edge e. That is, using the partition of ∂Ω induced by e, up to precision πM .

1: Set a list L = [ ] 2: set a list N with all neighbors of zero ⊲ we have an auxiliary routine to do that k 1 3: set h0 = [k,u mod π − O ] ⊲ the neighbor of inﬁnity ∞ 4: for all h N do ∈ 5: γ h ⊲ for the change of variable (cf. equation (3.4)) ← 6: w h[1][2] ← h[1][1] h[1][1] 7: set fn =1 z w t and fd =1 z′ w t ⊲ fn and fd are the numerator and − − − − the denominator, respectively cf. 4.4.3 § 8: C = z w and C = z′ w ⊲ these are the constants n − d − 9: append to L the list [fn, fd,Cn,Cd,γ] 10: end for

11: set u = h0[1][2] 12: if k =0 then 6 z u z′ u 13: set fn =1 πk−−1 t and fd =1 πk−−1 t −− − 14: T k 1 u set γ = 1 0 L 15: append to the list [fn, fd, 1, 1,γ] 16: end if 17: if k =0 then

18: set f =1 πzt and f =1 πz′t n − d − 0 T 19: set γ =( 1 0 )

20: append to L the list [fn, fd, 1, 1,γ] 21: end if 22: I = 1 and C =1 23: for l in L do

24: γ = l[5] and ﬁnd γrep, the representative of the edge given by γ and write γ = βγrepκ ⊲ use Algorithm 5 1 1 25: set h = l[1] κ− and h = l[2] κ− n h i d h i 26: deﬁne In to be the integral over O of hn ⊲ use Algorithm 11 ∞

146 B.2. Algorithms for the calculation of the integral

27: deﬁne Id to be the integral over O of hd ϕ(γ e ) ∞ l[3] rep 0 28: I I In ← × l[4] Id 29: end for 30: Return I

In the previous algorithm, we use an auxiliary routine to calculate the integral of a function f F . The implementation is an easy routine and we only give the algorithm without ∈ I further explanation.

Algorithm 11: Integrate over O ∞ Input: A function f B . A matrix γ a representative for the edges of G . The ∈ M rep N,M data structure GN,M of the quotient graph. The table T with the values of the integral

at the functions of BM . Output: The integral of f over O up to accuracy πM that is, ∞

f dµ . (B.15) × ϕ Z O∞ .

1: Set BM the pseudo basis for FI 2: Int= 1

3: set k to be the position in RN of γrep ⊲ The set RN is a list which is already stored in

the data structure of GN,M 4: run Algorithm 7 to set a list L of factors of f with multiplicities 5: for l in L do

6: set s to be the position in BM of l[2] 7: Int Int T[k][s]l[2] ← × 8: end for 9: Return Int

147 B. Algorithms for the table

Analysis for the algorithm Integrate over O ∞

A similar analysis as in Algorithm 7 shows that the worst case occurs when the function f B L d 2 factorizes as a product of all elements in M , then the list has 2 (M +3M) elements. So we only need to calculate the cost of the step 7. Each entry of the table T is an element of

the ring 1+πF 2 JπK and the exponent l[2] satisﬁes 1 6 l[2] 6 p 1 then using the powering q − algorithm (cf. [Coh93, Algorithm 1.2.1]) and one of the algorithms for fast multiplication allows to calculate the step 7 in O(log(p +1)M 2) operations, therefore the running time of the algorithm is O(M 4).

Analysis for the algorithm IntegrateOverEdge

We ignore the cost for the steps 1-21 since most of them are assignments or operations whose cost is O(1). Therefore we only consider the cost of the operations in the last loop.

In steps 26 and 27 we calculate the integral of the functions hn and hd, respectively, we know that the running time for Algorithm 11 is O(M 4). In step 28 we calculate three products so we need to consider the analysis separately. Note that the constants l[3] and l[4] do not depend on the size of M but on the path where we are integrating, so the cost ϕ(γ e ) l[3] rep 0 of calculating l[4] may be considered small (or even constant) compared with big values of M. The product I In can be carried out using fast multiplication in time O(M), × Id since they are polynomials of degree at most M (cf. Remark A.1.4). Therefore, adding up the costs for the steps 26, 27 and 28 we get that the algorithm takes time O(M 4).

Theorem B.2.1. The Tate parameter q can be calculated up to accuracy πM in time O(M 7).

Proof. In order to ﬁnd the Tate parameter we need to calculate the table an after the integral using Algorithms 6 and 10, respectively. The dominant cost of Algorithm 6 gives the running time for the calculation of q, which is O(M 7).

Remark B.2.2. Most of the computer algebra systems use a combination of the fast algorithms for multiplications or quotients of polynomials mentioned in Remark A.1.4 and they take M(n) 63.43n log n log log n + O(n log n). However in most of the cases we can ∈ not compute with M bigger than this constant, this is the reason we use for the analysis of the complexity the classical algorithm for multiplication.

148 C. Tables

C.1 Preliminaries

In this appendix we give some tables for the isogeny classes in characteristic 3 and 5 and conductor of degrees three and four in each case. In characteristic 2 there are already (using other methods) tables for conductor of small degree (cf. [Gek97],[Sch01],[Sch99]

and [Sch00]). Since applying the transformation T T + a for a in F× we can transform 7→ q any curve with bad reduction at the place T a to have bad reduction at the place T , − we only consider in the tables conductors which are divisible by T and the those that are primes.

Each table have three columns, the ﬁrst one is for the conductor, which is given by the factors of a polynomial N in F [T ], we omit the place, since we know that it appears q ∞ in the conductor with exponent 1. The second column is for the corresponding elliptic curve and the last one for the traces. All the traces have length 8, which correspond to the ﬁrst 8 prime polynomials relatively prime to N. We consider the list of primes ordered lexicographically.

Remark C.1.1. 1. As we already mentioned there are no elliptic curves with split multiplicative reduction at with conducto N and degree of N less than 2. ∞ ∞

2. Over F3 and F5 there are not elliptic curves with prime conductor of degree 3.

149 C. Tables

C.2 Table for degree 3 over F3

Table C.1: Isogeny classes for degree 3 over F3

Conductor Curve Trace T,T 2+2 Y 2=X3+(T 2+2)X2+(2T 8+T 6+2T 4)/(T 6+2) [0,0,2,2, 8, 8,8, 4] − − − T,T +1,T +2 Y 2=X3+(T 2+2T +2)X2+(2T 4+2T 3+T 2+T )/(T 6+2T 3+2) [2,2, 6, 4, 4,4,4, 4] − − − − Y 2=X3+(T 2+T +2)X2+(2T 4+T 3+T 2+2T )/(T 6+T 3+2) [2, 6,2, 4, 4,4, 4,4] − − − − Y 2=X3+(T 2+1)X2+(2T 4+T 2)/(T 6+1) [ 6,2,2, 4, 4, 4,4,4] − − − − T 2,T +1 Y 2+XY =X3+(2T 3+2)/T 9 [ 2, 2, 2,4,4, 8,4, 2] − − − − − Y 2=X3+(T 2+2T )X2+(2T 5+2T 4)/(T 3+2) [0,2,2, 2,4,4, 4, 8] − − − T 2,T +2 Y 2+XY =X3+(2T 3+1)/T 9 [ 2, 2,4, 2, 8,4,4,4] − − − − Y 2=X3+(T 2+T )X2+(2T 5+T 4)/(T 3+1) [0,2, 2,2,4,4,4, 4] − − T 2+2T +2,T Y 2=X3+(T 2+1)X2+(T 7+2T 6+2T 5)/(T 6+1) [ 1, 1,0,5,8, 7,8, 2] − − − − T 2+T +2,T Y 2=X3+(T 2+1)X2+(2T 7+T 6+T 5)/(T 6+1) [ 1, 1,0,5, 7,8, 7, 2] − − − − −

C.3 Table for degree 4 over F3

Table C.2: Isogeny classes for degree 3 over F3

Conductor Curve Trace T,T 3+2T 2+T +1 Y 2=X3+(T 2+2T +2)X2+(T 5+2T 4+T 3+T 2)/(T 6+2T 3+2) [ 2, 1, 1, 1,0, 1, 4, 8] − − − − − − − T,T 3+2T 2+1 Y 2+XY =X3+(T 7+2T 6+T 4)/(T 12+T 9+T 3+1) [ 3, 2, 2, 5,1, 5, 8,7] − − − − − − T 2,T +1,T +2 Y 2+XY =X3+(2T 2+1)/T 6 [ 2, 2, 2,4,4, 8,4,4] − − − − T 2,T 2+1 Y 2+XY =X3+(2T 2+2)/T 6 [ 2, 2, 2, 2, 2, 2, 2,4] − − − − − − − Y 2=X3+(T 2+T )X2+(2T 7+2T 5)/(T 3+1) [0,2,4, 4,4, 2, 2,8] − − − Y 2=X3+(T 2+2T )X2+(2T 7+2T 5)/(T 3+2) [2,0, 4,4, 2,4, 4,4] − − − T 4 Y 2+XY =X3+1/T 6 [ 2, 2, 5,1,1,4,4, 2] − − − − Y 2+XY =X3+2/T 6 [1,1, 2,1,1,7,7,4] − T 2,T 2+T +2 Y 2=X3+(T 2+2T )X2+(2T 2+2T +1)/(T 3+2) [ 1, 3,2, 5,1, 2, 1, 2] − − − − − − Y 2+XY =X3+(2T 6+2T 3+1)/T 9 [1,1, 2,1,7, 2, 5,10] − − − Y 2=X3+(T 2+T )X2+(2T 6+2T 5+T 4)/(T 3+1) [ 3, 1, 4, 1, 5, 8,1,2] − − − − − − Y 2=X3+(T 2+2T )X2+(2T 5+2T 4+T 3)/(T 3+2) [2,0,2, 2, 2, 2,8, 2] − − − − T 2,(T +2)2 Y 2=X3+1)X2+2T/(T 3+2) [ 2,4, 2, 2,4, 8, 8,4] − − − − − Y 2=X3+1)X2+(2T +1)/T 3 [ 2, 2,4, 2, 8,4,4,4] − − − − T 3,T +1 Y 2=X3+(T 2+2T )X2+(2T 4+T 3+2T 2)/(T 3+2) [ 3, 1, 4, 2, 2,1, 1,7] − − − − − − Y 2+XY =X3+(2T 6+T 3+2)/T 9 [1, 5,4, 2, 2,1,7, 5] − − − − Y 2+XY =X3+(T 3+1)/T 9 [1,4, 5, 2, 2,10, 2,4] − − − − Continued on next page

150 C.3. Table for degree 4 over F3

Conductor Curve Trace Y 2=X3+(T 2+2T )X2+(T 7+2T 6+T 5+T 4+2T 3+T 2)/(T 3+2) [3, 4, 1, 2,10, 2,2,4] − − − − T 2,T 2+2T +2 Y 2=X3+(T 2+T )X2+(2T 5+T 4+T 3)/(T 3+1) [0,2,2, 2, 2, 2,4,8] − − − Y 2=X3+(T 2+2T )X2+(T 6+2T 5+2T 4)/(T 3+2) [ 1, 3, 4, 1, 8, 5, 4, 2] − − − − − − − − Y 2+XY =X3+(T 6+2T 3+2)/T 9 [1,1, 2,1, 2,7,4, 8] − − − Y 2=X3+(T 2+T )X2+(T 2+2T +2)/(T 3+1) [ 3, 1,2, 5, 2,1, 8, 4] − − − − − − Y 2=X3+(T 2+T )X2+(T 8+2T 7+2T 6)/(T 3+1) [3, 1, 4,1,4,1, 8,2] − − − T 2,(T +1)2 Y 2=X3+1)X2+2T/(T 3+1) [ 2,4, 2, 2, 8,4,4, 2] − − − − − Y 2=X3+1)X2+(2T +2)/T 3 [ 2, 2, 2,4,4, 8,4, 2] − − − − − T,T 3+2T +2 Y 2=X3+(T 2+2T +2)X2+(T 4+2T 2+2T )/(T 6+2T 3+2) [ 2, 1,2, 1, 3, 7,1,1] − − − − − Y 2+XY =X3+(T 8+2T 6+2T 5)/(T 12+2T 9+2T 3+1) [ 2, 3, 2, 5,1,7, 5, 5] − − − − − − T 3,T +2 Y 2=X3+(T 2+T )X2+(T 4+T 3+T 2)/(T 3+1) [ 3, 1, 2, 4,1, 2, 2, 10] − − − − − − − Y 2+XY =X3+(T 6+T 3+1)/T 9 [1, 5, 2,4,1, 2, 2, 2] − − − − − Y 2+XY =X3+(T 3+2)/T 9 [1,4, 2, 5,10, 2,7, 2] − − − − Y 2=X3+(T 2+T )X2+(T 7+T 6+T 5+2T 4+2T 3+2T 2)/(T 3+1) [3, 4, 2, 1, 2,10,1,2] − − − − T,T +1,T 2+2T +2 Y 2=X3+(T 2+T +2)X2+(2T 6+2T 5+2T 4+T 2)/(T 6+T 3+2) [ 2, 4,0, 10,8,4, 4, 8] − − − − − Y 2+XY =X3+(2T 8+2T 6+2T 3+T 2)/(T 12+2T 9+2T 3+1) [0, 2, 2, 8,4,4,4,4] − − − Y 2=X3+(T 2+1)X2+2T 2(T +1)6(T 2+2T +2)/(T 6+1) [2,0, 4,2,8, 4,4, 8] − − − T,(T +2)3 Y 2=X3+(T 2+T )X2+2T 5/(T 4+2T 3+T +2) [3,2,1, 4, 2,1, 2, 1] − − − − Y 2+XY =X3+2T 6/(T 9+2) [1, 2, 5,4, 2,1,10,7] − − − Y 2+XY =X3+(T 3/(T 9+2) [1, 2,4, 5, 2,10,1, 2] − − − − T,(T +1)2,T +2 Y 2+XY =X3+(2T 2+T )/(T 6+2T 3+1) [ 2, 2, 2,4,4,4,4, 8] − − − − T,T 3+T 2+T +2 Y 2=X3+(T 2+T +2)X2+(2T 5+2T 4+2T 3+T 2)/(T 6+T 3+2) [ 1, 2, 1,0, 1, 4, 1,7] − − − − − − T,T 3+2T 2+2T +2 Y 2=X3+(T 2+2T +2)X2+(T 10+2T 9+2T 8+2T 7)/(T 6+2T 3+2) [1,2, 1, 1,0,8, 7,1] − − − T,T +2,T 2+1 Y 2=X3+(T 2+2T +2)X2+(2T 6+2T 5+T 4+2T 3+2T 2)/(T 6+2T 3+2) [ 2, 4,0, 10,8, 8,4, 4] − − − − − Y 2+XY =X3+(2T 8+2T 7+T 6+2T 5+2T 4)/(T 12+T 9+T 3+1) [0, 2, 2, 8,4,4,4,4] − − − Y 2=X3+(T 2+T +2)X2+(2T 10+2T 9+T 8+2T 7+2T 6)/(T 6+T 3+2) [2,0, 4,2,8, 8, 4,4] − − − T,T +1,T 2+T +2 Y 2+XY =X3+(2T 4+T 3+T )/(T 12+2T 9+2T 3+1) [0, 2, 2,4,4,4,4, 8] − − − Y 2=X3+(T 2+1)X2+(2T 8+T 7+T 6+T 5+T 3+2T 2+2T )/(T 6+1) [2,0, 4, 4, 4,8,4,10] − − − Y 2=X3+(T 2+2T +2)X2+(2T 8+2T 6+T 4+2T 3)/(T 6+2T 3+2) [2, 4,0, 4, 4, 8,4,2] − − − − T,T 3+2T +1 Y 2=X3+(T 2+T +2)X2+(T 4+2T 2+T )/(T 6+T 3+2) [ 1, 2,2, 3, 1, 7, 8, 7] − − − − − − − Y 2+XY =X3+(T 8+2T 6+T 5)/(T 12+T 9+T 3+1) [ 3, 2, 2,1, 5,7, 8,7] − − − − − T,T +2,T 2+T +2 Y 2=X3+(T 2+2T +2)X2+(2T 6+T 5+2T 4+T 2)/(T 6+2T 3+2) [ 2, 4,0,8, 10,4, 2, 4] − − − − − Y 2+XY =X3+(2T 8+2T 6+T 3+T 2)/(T 12+T 9+T 3+1) [0, 2, 2,4, 8,4, 8,4] − − − − Y 2=X3+(T 2+1)X2+2T 2(T +2)6(T 2+T +2)/(T 6+1) [2,0, 4,8,2, 4,10,4] − − T,(T +1)3 Y 2=X3+(T 2+2T )X2+T 5/(T 4+T 3+2T +2) [3,2, 4,1,1, 2, 10,4] − − − Y 2+XY =X3+(T 3/(T 9+1) [1, 2, 5,4,10, 2, 2, 5] − − − − − Y 2+XY =X3+(T 6/(T 9+1) [1, 2,4, 5,1, 2, 2,4] − − − − Y 2=X3+(T 2+2T )X2+T 8/(T 4+T 3+2T +2) [ 3,2, 1,4, 2,10,2,7] − − − T,T 3+T 2+2T +1 Y 2=X3+(T 2+T +2)X2+(T 10+T 9+2T 8+T 7)/(T 6+T 3+2) [2,1, 1,0, 1, 7,8,4] − − − T,T +1,T 2+1 Y 2=X3+(T 2+T +2)X2+(2T 6+T 5+T 4+T 3+2T 2)/(T 6+T 3+2) [ 2,0, 4,8, 10, 2, 4,4] − − − − − Continued on next page

151 C. Tables

Conductor Curve Trace Y 2+XY =X3+(2T 8+T 7+T 6+T 5+2T 4)/(T 12+2T 9+2T 3+1) [0, 2, 2,4, 8, 8,4,4] − − − − Y 2=X3+(T 2+2T +2)X2+(2T 10+T 9+T 8+T 7+2T 6)/(T 6+2T 3+2) [2, 4,0,8,2,10,4, 4] − − T,T +2,T 2+2T +2 Y 2+XY =X3+(2T 4+2T 3+2T )/(T 12+T 9+T 3+1) [0, 2, 2,4,4, 8,4,4] − − − Y 2=X3+(T 2+1)X2+(2T 8+2T 7+T 6+2T 5+2T 3+2T 2+T )/(T 6+1) [2,0, 4, 4, 4,2, 8,4] − − − − Y 2=X3+(T 2+T +2)X2+(2T 8+2T 6+T 4+T 3)/(T 6+T 3+2) [2, 4,0, 4, 4,10,8,4] − − − T,T +1,(T +2)2 Y 2+XY =X3+(2T 2+2T )/(T 6+T 3+1) [ 2, 2, 2,4,4,4, 8,4] − − − −

C.4 Table for degree 3 over F5

Table C.3: Isogeny classes for degree 3 over F5

Conductor Curve Trace T,T +2,T +3 Y 2=X3+(3T 4+3T 2+3)X+T 6+2T 5+T 4+2T 3+4T 2+2T +4 [ 2, 2, 6, 6,10,2,2, 6] − − − − − Y 2=X3+(3T 4+2T 3+3T +2)X+T 6+4T 5+2T 4+T 3+3T 2+3T +3 [0,0,2,2,2, 4,8,2] − Y 2=X3+(3T 4+3T 3+2T +2)X+T 6+T 5+2T 4+4T 3+3T 2+2T +3 [0,0,2,2, 10,8, 4,2] − − T,(T +1)2 Y 2=X3+(3T 4+T 3+4T +2)X+T 6+T 5+2T 4+3T 3+2T 2+4T +3 [3, 3,0, 4, 4, 7,2,8] − − − − Y 2=X3+(3T 4+T 3+4T +2)X+T 6+2T 5+T +2 [ 2,2,0,6,6, 2,2, 2] − − − Y 2=X3+(3T 4+2T 3+3T 2+2T +3)X+T 6+4T 5+T +4 [1,1, 4, 4, 4,1,6, 4] − − − − Y 2=X3+(3T 4+4T 3+2T 2+4T +3)X+T 6+T 5+4T 4+T 2+4T +4 [ 2, 2,2,2,2,10, 6,2] − − − T,(T +4)2 Y 2=X3+(3T 4+T 3+2T 2+T +3)X+T 6+2T 5+2T 4+T 3+3T +1 [2, 2, 2,2,2,6,2, 6] − − − Y 2=X3+(3T 4+3T 3+3T 2+3T +3)X+T 6+T 5+4T +4 [ 4,1,1, 4, 4,6, 4,6] − − − − Y 2=X3+(3T 4+4T 3+T +2)X+T 6+3T 5+4T +2 [0,2, 2,6,6, 8, 2,2] − − − Y 2=X3+(3T 4+4T 3+T +2)X+T 6+4T 5+2T 4+2T 3+2T 2+T +3 [0, 3,3, 4, 4,2,8,2] − − − T,T 2+2 Y 2=X3+(3T 4+T 2+2)X+T 6+T 2+3 [2, 2, 2,2, 6,2, 6,10] − − − − T,T 2+T +2 Y 2=X3+(3T 4+4T 3+2T 2+3T +2)X+T 6+T 5+3T 4+T 3+3T +3 [0,0, 3,3, 10,8, 1, 1] − − − − T,T +3,T +4 Y 2=X3+(3T 4+2T 3+T 2+4T +2)X+T 6+2T 5+3T 4+T 3+2T 2+T +2 [0,0, 4,2,8, 10,2,8] − − Y 2=X3+(3T 4+3T 3+4T +3)X+T 6+4T 4+3T 3+T 2+3T +4 [ 2, 2,2,10,2,10, 6,2] − − − Y 2=X3+(3T 4+4T 3+2T +3)X+T 6+T 5+3T 4+T +1 [0,0,8, 10, 4,2,2, 4] − − − T,T +1,T +2 Y 2=X3+(3T 4+T 3+3T +3)X+T 6+4T 5+3T 4+4T +1 [0,0,8, 10,2,8,2,2] − Y 2=X3+(3T 4+2T 3+T +3)X+T 6+4T 4+2T 3+T 2+2T +4 [ 2, 2,2,10, 6,2, 6, 6] − − − − − Y 2=X3+(3T 4+3T 3+T 2+T +2)X+T 6+3T 5+3T 4+4T 3+2T 2+4T +2 [0,0, 4,2,2, 4,2,2] − − T,T 2+4T +2 Y 2=X3+(3T 4+T 3+2T 2+2T +2)X+T 6+4T 5+3T 4+4T 3+2T +3 [3, 3,0,0, 10,8,8, 1] − − − T,T 2+3 Y 2=X3+(3T 4+4T 2+2)X+T 6+T 2+2 [ 2,2,2, 2, 6,2,10, 6] − − − − T,T +2,T +4 Y 2=X3+(3T 4+T 3+4T 2+3T +2)X+T 6+T 5+2T 4+2T 3+2T 2+3T +3 [0,0,2, 4,8,2, 4,2] − − Y 2=X3+(3T 4+2T 3+4T +3)X+T 6+3T 5+2T 4+3T +4 [0,0, 10,8, 4,2,8,2] − − Y 2=X3+(3T 4+4T 3+3T +3)X+T 6+T 4+T 3+T 2+4T +1 [ 2, 2,10,2,2, 6,2, 6] − − − − T,T 2+2T +3 Y 2=X3+(3T 4+3T 3+3T 2+4T +2)X+T 6+2T 5+2T 4+3T 3+T +2 [ 3,0,3,0,8, 10,8, 1] − − − T,T 2+3T +3 Y 2=X3+(3T 4+2T 3+3T 2+T +2)X+T 6+3T 5+2T 4+2T 3+4T +2 [0,3,0, 3,8, 10, 1, 1] − − − − Continued on next page

152 C.5. Table for non-primes of degree 4 over F5

Conductor Curve Trace T,T +1,T +3 Y 2=X3+(3T 4+T 3+2T +3)X+T 6+T 4+4T 3+T 2+T +1 [ 2, 2,10,2, 6, 6,10,2] − − − − Y 2=X3+(3T 4+3T 3+T +3)X+T 6+2T 5+2T 4+2T +4 [0,0, 10,8,2,2,2, 4] − − Y 2=X3+(3T 4+4T 3+4T 2+2T +2)X+T 6+4T 5+2T 4+3T 3+2T 2+2T +3 [0,0,2, 4,2,2, 10,8] − − T,T +1,T +4 Y 2=X3+(3T 4+2T 2+3)X+T 6+T 5+4T 4+4T 3+4T 2+T +1 [ 2, 2, 6, 6, 6,2,2,10] − − − − − Y 2=X3+(3T 4+T 3+T +2)X+T 6+2T 5+3T 4+2T 3+3T 2+4T +2 [0,0,2,2,2,8,8, 10] − T,(T +3)2 Y 2=X3+(3T 4+T 3+2T 2+4T +3)X+T 6+2T 5+3T +1 [1, 4,1, 4, 4,1, 4, 4] − − − − − Y 2=X3+(3T 4+2T 3+3T 2+3T +3)X+T 6+3T 5+T 4+T 2+2T +1 [ 2,2, 2,2,2, 2,2,2] − − − Y 2=X3+(3T 4+3T 3+3T +2)X+T 6+T 5+3T +3 [ 2,0,2,6,6,4, 2, 2] − − − Y 2=X3+(3T 4+3T 3+3T +2)X+T 6+3T 5+3T 4+T 3+2T 2+2T +2 [3,0, 3, 4, 4, 1,8,8] − − − − T,(T +2)2 Y 2=X3+(3T 4+2T 3+2T +2)X+T 6+4T 5+2T +3 [2,0, 2,6,6,4,2,2] − Y 2=X3+(3T 4+2T 3+2T +2)X+T 6+2T 5+3T 4+4T 3+2T 2+3T +2 [ 3,0,3, 4, 4, 1,2,2] − − − − Y 2=X3+(3T 4+3T 3+3T 2+2T +3)X+T 6+T 5+3T 4+2T 3+4T +4 [ 2,2, 2,2,2, 2, 6, 6] − − − − − Y 2=X3+(3T 4+4T 3+2T 2+T +3)X+T 6+3T 5+2T +1 [1, 4,1, 4, 4,1,6,6] − − − T 2,T +1 Y 2=X3+(3T 4)X+T 6+2T 5 [ 4,1,1, 4,1,1,6, 4] − − − Y 2=X3+(3T 4+T 3)X+T 6+2T 4 [0, 3,3,8, 1, 7,2, 4] − − − − Y 2=X3+(3T 4+T 3)X+T 6+4T 5 [0,2, 2, 2,4, 2,2,6] − − − Y 2=X3+(3T 4+3T 3+3T 2)X+T 6+4T 4+T 3 [2, 2, 2,2, 2,10, 6,2] − − − − T 2,T +2 Y 2=X3+(3T 4)X+T 6+4T 5 [1,1, 4,1, 4, 4,6,6] − − − Y 2=X3+(3T 4+T 3+2T 2)X+T 6+T 4+3T 3 [ 2, 2,2, 2,2,2,6, 6] − − − − Y 2=X3+(3T 4+2T 3)X+T 6+3T 4 [ 3,3,0, 1,8,8,2,2] − − Y 2=X3+(3T 4+2T 3)X+T 6+3T 5 [2, 2,0,4, 2, 2, 8,2] − − − − T 2,T +3 Y 2=X3+(3T 4)X+T 6+T 5 [ 4,1,1,1, 4, 4, 4,1] − − − − Y 2=X3+(3T 4+3T 3)X+T 6+3T 4 [0,3, 3, 1,8, 4, 4, 1] − − − − − Y 2=X3+(3T 4+3T 3)X+T 6+2T 5 [0, 2,2,4, 2,6,6,4] − − Y 2=X3+(3T 4+4T 3+2T 2)X+T 6+T 4+2T 3 [2, 2, 2, 2,2,2,2, 2] − − − − T 2,T +4 Y 2=X3+(3T 4)X+T 6+3T 5 [1,1, 4, 4,1,6,1,6] − − Y 2=X3+(3T 4+2T 3+3T 2)X+T 6+4T 4+4T 3 [ 2, 2,2,2, 2, 6, 2,6] − − − − − Y 2=X3+(3T 4+4T 3)X+T 6+2T 4 [3, 3,0,8, 1,2, 1,2] − − − Y 2=X3+(3T 4+4T 3)X+T 6+T 5 [ 2,2,0, 2,4,2,4, 8] − − −

C.5 Table for non-primes of degree 4 over F5

Table C.4: Isogeny classes for non-primes of degree 4 over F5

Conductor Curve Trace T,T +1,T 2+4T +2 Y 2=X3+(3T 4+3T 2+T +2)X+T 6+2T 5+2T 4+T 3+4T +2 [0,3,3, 1,2,2,2, 7] − − Y 2=X3+(3T 4+2T 3+2)X+T 6+T 5+T 4+T 3+2T 2+3T +2 [ 2, 3, 1,7, 4, 8,2, 3] − − − − − − Y 2=X3+(3T 4+2T 3+T 2+2T +2)X+T 6+T 5+4T 4+2T 3+4T 2+2T +3 [ 2, 2,0,8,6,2, 8,4] − − − Continued on next page

153 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+4T 3+T 2+3)X+T 6+2T 5+2T 4+3T 3+2T 2+4 [ 4, 1, 3, 7, 6,2, 2,7] − − − − − − T,T +4,T 2+2T +3 Y 2=X3+(3T 4+T 3+T 2+T +2)X+T 6+T 4+T 3+2T 2+4T +2 [ 2, 2, 2, 6,2, 2,2, 6] − − − − − − Y 2=X3+(3T 4+T 3+T 2+4T +3)X+T 6+3T 5+2T 4+2T 3+2T 2+2T +1 [ 1, 2, 3, 2, 2, 2,1,1] − − − − − − Y 2=X3+(3T 4+2T 3+4T 2+T +2)X+T 6+T 5+4T 2+4T +2 [2,1,3,1,1, 5, 2, 2] − − − T,(T +2)2,T +3 Y 2=X3+(3T 4+4T 2+3T +2)X+T 6+2T 4+4T 3+T +3 [ 3, 1,6, 6,3, 8, 4,3] − − − − − Y 2=X3+(3T 4+4T 2+3T +2)X+T 6+4T 5+4T 3+2T 2+2T +2 [2,4,6, 6, 2,2, 4, 2] − − − − Y 2=X3+(3T 4+T 3+4T +3)X+T 6+4T 4+2T 3+T 2+4 [2,2,2,2, 2, 2, 2, 6] − − − − Y 2=X3+(3T 4+T 3+4T +3)X+T 6+2T 5+3T 4+2T 3+2T 2+2T +4 [ 3, 3,2,2, 7,8,8, 1] − − − − Y 2=X3+(3T 4+T 3+T 2+3T +2)X+T 6+T 5+2T 4+4T 3+T 2+2T +3 [4,0, 6, 6,6, 4,4, 6] − − − − Y 2=X3+(3T 4+3T 3+T 2+2)X+T 6+3T 5+2T 3+4T 2+2 [1,3, 6,6,3, 4, 8,3] − − − Y 2=X3+(3T 4+3T 3+T 2+2)X+T 6+4T 5+T 4+3T 3+3T +3 [ 4, 2, 6,6, 2, 4,2, 2] − − − − − − Y 2=X3+(3T 4+3T 3+4T 2+T +2)X+T 6+T 5+2T 4+2T 3+4T 2+3T +2 [0, 4, 6, 6, 6,4, 4, 6] − − − − − − Y 2=X3+(3T 4+4T 3+2T 2+T +3)X+T 6+T 5+2T 4+T 3+3T 2+T +4 [1,1,6,6,1, 4, 4, 9] − − − T,T 3+4T 2+T +2 Y 2=X3+(3T 4+T 3+T 2+T +3)X+T 6+3T 5+2T 2+3T +4 [0,0,0,3, 1, 1, 4, 7] − − − − T,T +2,(T +4)2 Y 2=X3+,3T 4+3T 2+2T +2)X+T 6+2T 5+4T 4+4T 3+4T 2+3T +2 [2,2, 2, 2, 2, 6,6, 10] − − − − − Y 2=X3+(3T 4+3T 2+2T +2)X+T 6+4T 5+3T 4+T 3+T 2+2T +3 [ 3, 3, 7,8,8, 1, 4, 10] − − − − − − Y 2=X3+(3T 4+4T 2+3)X+T 6+3T 5+2T 4+3T 2+1 [0,4,6, 4,4, 6,4,6] − − Y 2=X3+(3T 4+2T 3+T 2+T +3)X+T 6+2T 3+T 2+2T +4 [3,1,3, 4, 8,3, 8, 6] − − − − Y 2=X3+(3T 4+2T 3+T 2+T +3)X+T 6+T 5+T 4+T 3+2T 2+3T +1 [ 2, 4, 2, 4,2, 2, 8,4] − − − − − − Y 2=X3+(3T 4+2T 3+4T 2+4T +2)X+T 6+3T 5+2T 4+3T 3+4T 2+4T +3 [ 4,0, 6,4, 4, 6, 4,6] − − − − − Y 2=X3+(3T 4+3T 3+3T 2+3T +3)X+T 6+3T 5+2T 4+2T 3+2T 2+T +4 [1,1,1, 4, 4, 9, 4,6] − − − − Y 2=X3+(3T 4+4T 3+T 2+2)X+T 6+T 5+4T 3+T 2+3 [4,2, 2,2, 4, 2, 2,4] − − − − Y 2=X3+(3T 4+4T 3+T 2+2)X+T 6+2T 5+2T 4+2T 2+3 [ 1, 3,3, 8, 4,3,8, 6] − − − − − T,T +1,T 2+2 Y 2=X3+(3T 4+2T 3+3T 2+4T +2)X+T 6+2T 5+T 4+T 3+4T +3 [1,2,3, 7, 8,9, 4,7] − − − Y 2=X3+(3T 4+2T 3+4T 2+3)X+T 6+T 5+3T 4+1 [ 3, 4, 1,3,2, 3, 6, 7] − − − − − − Y 2=X3+(3T 4+3T 3+3T 2+4T +3)X+T 6+3T 5+3T 4+3T +4 [ 3,0, 3, 1,2, 1,2, 1] − − − − − Y 2=X3+(3T 4+4T 3+3T 2+3T +3)X+T 6+2T 5+T 4+T 2+4T +1 [0,2,2,6,2, 6,6,8] − T,T 3+2T 2+2T +2 Y 2=X3+(3T 4+4T 3+T 2+4T +3)X+T 6+3T 5+3T 4+3T +4 [3,0,0,0, 1, 1, 1,2] − − − T,T 3+3T 2+2T +2 Y 2=X3+(3T 4+4T 2+4T +3)X+T 6+3T 5+2T 4+3T +4 [ 3,3,0,3, 1,8,2, 4] − − − Y 2=X3+(3T 4+T 3+3T 2+2T +3)X+T 6+3T 5+3T 4+2T 3+T +1 [ 3, 3,0, 1,9, 4,2, 8] − − − − − Y 2=X3+(3T 4+T 3+3T 2+3T +3)X+T 6+3T 5+3T 4+3T 2+4T +1 [0, 2, 1, 4, 3,4,3,2] − − − − Y 2=X3+(3T 4+4T 3+3T 2+2)X+T 6+3T 5+T 4+4T 3+3T 2+3 [ 2,0,3,4,3, 2, 9, 8] − − − − Y 2=X3+(3T 4+4T 3+3T 2+4T +3)X+T 6+4T 5+3T 4+4T 3+2T 2+3T +4 [ 3,1, 4, 1, 3, 8, 6,8] − − − − − − Y 2=X3+(3T 4+4T 3+4T 2+3T +2)X+T 6+2T 5+T 4+T 2+2T +2 [ 1,1, 2, 3, 1, 2,6, 4] − − − − − − T,T +3,T 2+3T +4 Y 2=X3+(3T 4+T 2+2T +2)X+T 6+2T 5+T 4+T 2+3T +2 [ 2,0, 4, 2, 8, 8,4, 2] − − − − − − Y 2=X3+(3T 4+T 3+3T 2+3T +3)X+T 6+T 5+T 4+2T 3+4T +1 [0, 2, 4, 8,4, 2, 8,4] − − − − − Y 2=X3+(3T 4+2T 3+3T +3)X+T 6+3T 5+T 4+4T 3+4T 2+3T +4 [2, 2,2, 2, 2, 6,6,6] − − − − Y 2=X3+(3T 4+2T 3+3T 2+4T +3)X+T 6+T 5+3T 3+T 2+4T +1 [ 4, 2,0,4, 8, 2,4, 8] − − − − − Y 2=X3+(3T 4+2T 3+4T 2+2)X+T 6+T 5+4T 4+4T 2+2 [4,0,2, 8, 2,4, 2,4] − − − Y 2=X3+(3T 4+3T 3+4T 2+T +3)X+T 6+4T 5+2T 4+2T 3+4T 2+2T +4 [2,2, 2,6, 6,6, 6, 2] − − − − Y 2=X3+(3T 4+4T 3+2T 2+4T +2)X+T 6+2T 5+2T 3+4T 2+4T +3 [ 2, 4,0, 2,4,4, 8, 2] − − − − − Continued on next page

154 C.5. Table for non-primes of degree 4 over F5

Conductor Curve Trace Y 2=X3+(3T 4+4T 3+3T 2+4T +2)X+T 6+2T 5+4T 4+3T 3+4T +3 [0,4,2,4, 2, 8, 2, 8] − − − − Y 2=X3+(3T 4+4T 3+4T 2+2T +3)X+T 6+T 4+3T 3+2T 2+4T +4 [ 2,2,2, 6,6, 2, 2, 6] − − − − − T,T +1,T 2+T +2 Y 2=X3+(3T 4+T 3+T 2+2T +2)X+T 6+3T 5+2T 4+2T +3 [ 1, 3, 2,1,1,3, 2, 2] − − − − − Y 2=X3+(3T 4+T 3+T 2+4T +3)X+T 6+3T 5+2T 4+3T 3+2T 2+2T +1 [ 2, 3, 1, 2, 2,3,1,1] − − − − − Y 2=X3+(3T 4+T 3+4T 2+T +2)X+T 6+3T 5+3T 4+T 3+T 2+T +3 [2,2,2,2, 6, 6,2, 6] − − − T,T +2,T 2+T +1 Y 2=X3+(3T 4+2T 2+3T +2)X+T 6+T 4+2T 3+2T 2+4T +3 [ 2, 2,0, 8,4, 6, 8,2] − − − − − Y 2=X3+(3T 4+3T 2+3T +3)X+T 6+4T 4+4T 3+T +4 [ 3, 2, 1, 4, 3,1,2, 8] − − − − − − Y 2=X3+(3T 4+T 3+3T 2+T +3)X+T 6+3T 5+3T 4+3T +1 [ 1, 4, 3, 8,7,3, 2,2] − − − − − Y 2=X3+(3T 4+4T 3+4T 2+T +2)X+T 6+3T 5+3T 4+T 3+4T 2+4T +2 [3,0,3, 4, 7, 1,2,2] − − − T,T +4,T 2+3 Y 2=X3+(3T 4+4T 2+3T +3)X+T 6+3T 2+4T +1 [ 1, 3, 4,3, 2, 6, 3, 7] − − − − − − − Y 2=X3+(3T 4+T 3+3T 2+2T +3)X+T 6+3T 5+2T 2+3T +4 [ 3, 3,0, 1,2,2, 1, 1] − − − − − Y 2=X3+(3T 4+2T 3+3T 2+T +3)X+T 6+T 4+3T 3+2T 2+2T +4 [2,0,2,6, 8,6, 6,8] − − Y 2=X3+(3T 4+4T 3+2T 2+2T +2)X+T 6+2T 5+3T 3+2T 2+T +2 [ 3, 1, 2, 7,2, 4,9,7] − − − − − T,T 3+4T +2 Y 2=X3+(3T 4+3T 3+3)X+T 6+4T 5+T 4+4T 3+1 [ 4, 1,0, 2,8,2, 2, 2] − − − − − T,T +2,T 2+4T +1 Y 2=X3+(3T 4+T 3+T 2+4T +2)X+T 6+3T 5+2T 2+T +2 [ 3,0, 3, 7, 4,2,2, 1] − − − − − Y 2=X3+(3T 4+2T 3+3T 2+3)X+T 6+T 5+4T 3+T 2+4 [ 1, 2, 3, 3, 4, 4,2,1] − − − − − − Y 2=X3+(3T 4+3T 3+3T 2+2)X+T 6+2T 5+4T 4+T 3+2T 2+2 [0,2,2,4, 8,6, 8, 6] − − − Y 2=X3+(3T 4+4T 3+3T 2+4T +3)X+T 6+2T 5+3T 2+2T +1 [ 3, 4, 1,7, 8, 6, 2,3] − − − − − − T,T +1,T +3,T +4 Y 2=X3+(3T 4+4T 2+3T +3)X+T 6+T 5+4T 4+T 3+4T 2+4T +4 [2, 2, 6, 2, 6,2, 2,2] − − − − − Y 2=X3+(3T 4+T 3+3T 2+4T +2)X+T 6+T 4+4T +3 [2, 2,2, 6, 2,2,2, 6] − − − − Y 2=X3+(3T 4+2T 3+4T 2+3)X+T 6+3T 5+4T 4+T 3+2T 2+1 [2, 6, 2,2,2, 6, 2, 2] − − − − − Y 2=X3+(3T 4+3T 3+2T 2+2T +3)X+T 6+T 4+T 3+4T 2+4 [2,2,2,2, 2, 2, 6, 2] − − − − Y 2=X3+(3T 4+4T 3+2T 2+T +2)X+T 6+2T 5+2T 4+T 3+2T 2+2T +3 [ 2,2, 2, 2,2, 2,2,2] − − − − T,T 3+4T 2+4T +2 Y 2=X3+(3T 4+2T 2+4T +2)X+T 6+T 4+2T 3+2T 2+4T +2 [ 4,0, 3,2, 2,3,5, 4] − − − − Y 2=X3+(3T 4+T 3+3T 2+T +3)X+T 6+3T 5+3T 4+4T 3+2T 2+4T +4 [ 1,1, 4, 3, 8, 3, 1, 2] − − − − − − − Y 2=X3+(3T 4+T 3+4T 2+3)X+T 6+3T 5+3T 2+3T +4 [3,3,0, 3,8, 1, 1, 4] − − − − Y 2=X3+(3T 4+2T 3+3T 2+4T +3)X+T 6+T 5+3T 4+3T 2+2T +1 [ 4, 2, 1,0,4, 3, 7,4] − − − − − Y 2=X3+(3T 4+3T 3+T 2+4T +2)X+T 6+4T 5+3T 4+3T 2+4T +3 [3, 1,2,1, 2, 1,7,2] − − − Y 2=X3+(3T 4+3T 3+3T 2+4T +3)X+T 6+4T 5+3T 3+3T 2+2T +1 [ 1, 3,0, 3, 4,9,7, 6] − − − − − T,T +2,T 2+3T +4 Y 2=X3+(3T 4+T 3+3T 2+3T +2)X+T 6+3T 5+3T 4+2T +2 [ 2, 3, 1,4, 5,1, 2, 5] − − − − − − Y 2=X3+(3T 4+T 3+3T 2+4T +3)X+T 6+2T 5+4T 4+T 2+2T +4 [ 2, 2, 2,6,6,2, 2, 2] − − − − − Y 2=X3+(3T 4+3T 3+2T 2+T +2)X+T 6+4T 5+4T 2+T +3 [1,3,2, 5,4, 2, 5, 2] − − − − T,T +3,(T +4)2 Y 2=X3+,3T 4+4T 2+3)X+T 6+4T 5+2T 3+4T 2+4 [0,4, 4,6,4, 6, 6, 4] − − − − Y 2=X3+(3T 4+4T 2+T +2)X+T 6+2T 4+4T 3+4T 2+T +3 [ 4, 2,2, 2, 4, 2, 6, 8] − − − − − − − Y 2=X3+(3T 4+4T 2+T +2)X+T 6+4T 4+4T 2+4T +2 [1,3, 8,3, 4,3, 6, 8] − − − − Y 2=X3+(3T 4+T 3+T 2+2T +3)X+T 6+3T 5+2T 4+T 2+4T +4 [ 2, 4, 4, 2,2, 2,6, 2] − − − − − − Y 2=X3+(3T 4+T 3+T 2+2T +3)X+T 6+3T 5+4T 4+3T 3+4 [3,1, 4,3, 8,3,6,8] − − Y 2=X3+(3T 4+T 3+T 2+3T +2)X+T 6+4T 5+3T 3+T 2+3T +3 [4,0,4, 6, 4,6, 6,4] − − − Y 2=X3+(3T 4+3T 3+2T 2+2)X+T 6+T 5+4T 4+T 2+T +2 [ 2, 2, 2, 2, 2, 2,2,6] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+2)X+T 6+4T 5+2T 2+3 [3,3,8, 7,8, 7,2, 4] − − − Continued on next page

155 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+3T 3+3T 2+3T +3)X+T 6+4T 5+3T 4+3T 3+3T 2+2T +4 [1,1, 4,1, 4,1,6, 4] − − − T,T 3+2T 2+3 Y 2=X3+(3T 4+3T +3)X+T 6+4T 3+T 2+4T +1 [ 4,0, 1, 2,2,8, 2, 1] − − − − − T,T +1,T 2+2T +3 Y 2=X3+(3T 4+3T +2)X+T 6+4T 5+T 4+3T 3+3T 2+3T +3 [3,2,1, 4,7, 8,9, 7] − − − Y 2=X3+(3T 4+T 2+4T +3)X+T 6+3T 4+2T 3+3T 2+3T +4 [ 1, 4, 3, 6, 7,2, 3,3] − − − − − − Y 2=X3+(3T 4+3T 3+4T 2+3T +2)X+T 6+4T 5+T 4+4T 3+2T 2+3T +3 [2,2,0,6,8,2, 6,6] − Y 2=X3+(3T 4+4T 3+2T 2+2)X+T 6+3T 5+3T 4+2T 3+3T 2+2 [ 3,0, 3,2, 1,2, 1, 1] − − − − − T,T +4,T 2+T +2 Y 2=X3+(3T 4+3T 2+4T +2)X+T 6+3T 5+2T 4+4T 3+T +2 [3,3,0, 1,2, 1, 1, 1] − − − − Y 2=X3+(3T 4+T 3+T 2+3)X+T 6+3T 5+2T 4+2T 3+2T 2+4 [ 3, 1, 4, 7, 6,3, 3,3] − − − − − − Y 2=X3+(3T 4+3T 3+2)X+T 6+4T 5+T 4+4T 3+2T 2+2T +2 [ 1, 3, 2,7, 4,1,9, 7] − − − − − Y 2=X3+(3T 4+3T 3+T 2+3T +2)X+T 6+4T 5+4T 4+3T 3+4T 2+3T +3 [0, 2, 2,8,6, 6, 6,6] − − − − T,T 3+T 2+T +3 Y 2=X3+(3T 4+4T 3+T 2+4T +3)X+T 6+2T 5+2T 2+2T +4 [3,0,0,0, 1, 1, 1,2] − − − T,T +2,(T +3)2 Y 2=X3+,3T 4+4T 2+2T +2)X+T 6+2T 4+T 3+4T +3 [ 1, 3,6, 6,3, 8,8, 6] − − − − − Y 2=X3+(3T 4+4T 2+2T +2)X+T 6+T 5+T 3+2T 2+3T +2 [4,2,6, 6, 2, 8, 2,4] − − − − Y 2=X3+(3T 4+T 3+2T 2+4T +3)X+T 6+4T 5+2T 4+4T 3+3T 2+4T +4 [1,1,6,6,1, 4, 4,6] − − Y 2=X3+(3T 4+2T 3+T 2+2)X+T 6+T 5+T 4+2T 3+2T +3 [ 2, 4, 6,6, 2, 2, 8,4] − − − − − − Y 2=X3+(3T 4+2T 3+T 2+2)X+T 6+2T 5+3T 3+4T 2+2 [3,1, 6,6,3,8, 8, 6] − − − Y 2=X3+(3T 4+2T 3+4T 2+4T +2)X+T 6+4T 5+2T 4+3T 3+4T 2+2T +2 [ 4,0, 6, 6,6,4, 4,6] − − − − Y 2=X3+(3T 4+4T 3+T +3)X+T 6+4T 4+3T 3+T 2+4 [2,2,2,2, 2,6,6, 10] − − Y 2=X3+(3T 4+4T 3+T +3)X+T 6+3T 5+3T 4+3T 3+2T 2+3T +4 [ 3, 3,2,2, 7, 4, 4, 10] − − − − − − Y 2=X3+(3T 4+4T 3+T 2+2T +2)X+T 6+4T 5+2T 4+T 3+T 2+3T +3 [0,4, 6, 6, 6, 4,4,6] − − − − T,T +1,T 2+3T +3 Y 2=X3+(3T 4+3T 3+4T 2+4T +2)X+T 6+4T 5+4T 2+T +2 [3,1,2,1,1,3, 2,4] − Y 2=X3+(3T 4+4T 3+T 2+T +3)X+T 6+2T 5+2T 4+3T 3+2T 2+3T +1 [ 3, 2, 1, 2, 2,3, 5, 5] − − − − − − − Y 2=X3+(3T 4+4T 3+T 2+4T +2)X+T 6+T 4+4T 3+2T 2+T +2 [ 2, 2, 2, 6,2, 6, 2,6] − − − − − − T,(T +1)2,T +3 Y 2=X3+(3T 4+3T 2+3T +2)X+T 6+2T 5+3T 4+2T 2+2T +3 [2,2, 2, 2,2,2, 2,6] − − − Y 2=X3+(3T 4+3T 2+3T +2)X+T 6+T 5+3T 4+4T 3+T 2+3T +3 [ 3, 3, 7,8,2,2, 7, 4] − − − − − Y 2=X3+(3T 4+4T 2+3)X+T 6+2T 5+2T 4+3T 2+1 [4,0,6, 4, 6, 6, 6, 4] − − − − − Y 2=X3+(3T 4+T 3+T 2+2)X+T 6+3T 5+2T 4+2T 2+3 [ 3, 1,3, 8,6, 6,3, 8] − − − − − Y 2=X3+(3T 4+T 3+T 2+2)X+T 6+4T 5+T 3+T 2+3 [2,4, 2,2,6, 6, 2, 8] − − − − Y 2=X3+(3T 4+2T 3+3T 2+2T +3)X+T 6+2T 5+2T 4+3T 3+2T 2+4T +4 [1,1,1, 4,6,6,1, 4] − − Y 2=X3+(3T 4+3T 3+T 2+4T +3)X+T 6+3T 3+T 2+3T +4 [1,3,3, 4, 6,6,3,8] − − Y 2=X3+(3T 4+3T 3+T 2+4T +3)X+T 6+4T 5+T 4+4T 3+2T 2+2T +1 [ 4, 2, 2, 4, 6,6, 2, 2] − − − − − − − Y 2=X3+(3T 4+3T 3+4T 2+T +2)X+T 6+2T 5+2T 4+2T 3+4T 2+T +3 [0, 4, 6,4, 6, 6,6,4] − − − − T,T 3+2T 2+2T +3 Y 2=X3+(3T 4+4T 2+T +3)X+T 6+2T 5+2T 4+2T +4 [3,0,3, 3, 1,8, 1, 7] − − − − Y 2=X3+(3T 4+T 3+3T 2+2)X+T 6+2T 5+T 4+T 3+3T 2+3 [4,3,0, 2,3, 2,5,9] − − Y 2=X3+(3T 4+T 3+3T 2+T +3)X+T 6+T 5+3T 4+T 3+2T 2+2T +4 [ 1, 4,1, 3, 3, 8, 1,3] − − − − − − Y 2=X3+(3T 4+T 3+4T 2+2T +2)X+T 6+3T 5+T 4+T 2+3T +2 [ 3, 2,1, 1, 1, 2,7,1] − − − − − Y 2=X3+(3T 4+4T 3+3T 2+2T +3)X+T 6+2T 5+3T 4+3T 2+T +1 [ 4, 1, 2,0, 3,4, 7, 9] − − − − − − Y 2=X3+(3T 4+4T 3+3T 2+3T +3)X+T 6+2T 5+3T 4+3T 3+4T +1 [ 1,0, 3, 3,9, 4,7, 1] − − − − − T,T 3+3T 2+2T +3 Y 2=X3+(3T 4+T 3+T 2+T +3)X+T 6+2T 5+3T 4+2T +4 [0,0,0,3, 1, 1, 4,8] − − − T,T +4,T 2+2 Y 2=X3+(3T 4+T 3+3T 2+2T +3)X+T 6+3T 5+T 4+T 2+T +1 [2,2,0,6, 8, 8, 6,4] − − − Y 2=X3+(3T 4+2T 3+3T 2+T +3)X+T 6+2T 5+3T 4+2T +4 [ 3,0, 3, 1,2, 4, 1, 7] − − − − − − Continued on next page

156 C.5. Table for non-primes of degree 4 over F5

Conductor Curve Trace Y 2=X3+(3T 4+3T 3+3T 2+T +2)X+T 6+3T 5+T 4+4T 3+T +3 [3,2,1, 7,2, 4,1, 3] − − − Y 2=X3+(3T 4+3T 3+4T 2+3)X+T 6+4T 5+3T 4+1 [ 1, 4, 3,3, 2, 8,3,7] − − − − − T,T +1,T 2+3 Y 2=X3+(3T 4+4T 2+2T +3)X+T 6+3T 2+T +1 [ 4, 3, 1,3,2,3, 8,7] − − − − Y 2=X3+(3T 4+T 3+2T 2+3T +2)X+T 6+3T 5+2T 3+2T 2+4T +2 [ 2, 1, 3, 7, 8,1, 4, 3] − − − − − − − Y 2=X3+(3T 4+3T 3+3T 2+4T +3)X+T 6+T 4+2T 3+2T 2+3T +4 [2,0,2,6,2, 6, 8,4] − − Y 2=X3+(3T 4+4T 3+3T 2+3T +3)X+T 6+2T 5+2T 2+2T +4 [0, 3, 3, 1,2, 1, 4, 7] − − − − − − T,T +3,T 2+4T +1 Y 2=X3+(3T 4+2T 2+2T +2)X+T 6+T 4+3T 3+2T 2+T +3 [0, 2, 2, 8,4,6,8,6] − − − Y 2=X3+(3T 4+3T 2+2T +3)X+T 6+4T 4+T 3+4T +4 [ 1, 2, 3, 4, 3, 4,7, 7] − − − − − − − Y 2=X3+(3T 4+T 3+4T 2+4T +2)X+T 6+2T 5+3T 4+4T 3+4T 2+T +2 [3,0,3, 4, 7,2, 1, 1] − − − − Y 2=X3+(3T 4+4T 3+3T 2+4T +3)X+T 6+2T 5+3T 4+2T +1 [ 3, 4, 1, 8,7, 6, 7,3] − − − − − − T,T +4,T 2+4T +2 Y 2=X3+(3T 4+4T 3+T 2+T +3)X+T 6+2T 5+2T 4+2T 3+2T 2+3T +1 [ 1, 3, 2, 2, 2, 2, 5, 5] − − − − − − − − Y 2=X3+(3T 4+4T 3+T 2+3T +2)X+T 6+2T 5+2T 4+3T +3 [ 2, 3, 1,1,1, 5,4, 2] − − − − − Y 2=X3+(3T 4+4T 3+4T 2+4T +2)X+T 6+2T 5+3T 4+4T 3+T 2+4T +3 [2,2,2,2, 6, 2,6, 2] − − − T,T +2,T 2+2T +4 Y 2=X3+(3T 4+T 2+3T +2)X+T 6+3T 5+T 4+T 2+2T +2 [ 4,0, 2, 2, 8,4, 2, 2] − − − − − − Y 2=X3+(3T 4+T 3+2T 2+T +2)X+T 6+3T 5+3T 3+4T 2+T +3 [0, 4, 2, 2,4, 8, 2,6] − − − − − Y 2=X3+(3T 4+T 3+3T 2+T +2)X+T 6+3T 5+4T 4+2T 3+T +3 [2,4,0,4, 2,4,6, 2] − − Y 2=X3+(3T 4+T 3+4T 2+3T +3)X+T 6+T 4+2T 3+2T 2+T +4 [2,2, 2, 6,6,6, 6,2] − − − Y 2=X3+(3T 4+2T 3+4T 2+4T +3)X+T 6+T 5+2T 4+3T 3+4T 2+3T +4 [ 2,2,2,6, 6, 2,2, 6] − − − − Y 2=X3+(3T 4+3T 3+2T +3)X+T 6+T 5+T 3+T +4 [2, 2,2, 2, 2, 6,2,2] − − − − Y 2=X3+(3T 4+3T 3+3T 2+T +3)X+T 6+4T 5+2T 3+T 2+T +1 [0, 2, 4,4, 8, 2, 2,6] − − − − − Y 2=X3+(3T 4+3T 3+4T 2+2)X+T 6+4T 5+4T 4+4T 2+2 [2,0,4, 8, 2, 8, 2, 2] − − − − − Y 2=X3+(3T 4+4T 3+3T 2+2T +3)X+T 6+4T 5+T 4+3T 3+T +1 [ 4, 2,0, 8,4, 2,6, 2] − − − − − T,T 3+4T +3 Y 2=X3+(3T 4+2T 3+3)X+T 6+T 5+T 4+T 3+1 [ 2,0, 1, 4,8,2, 9,4] − − − − T,T 3+T 2+4T +3 Y 2=X3+(3T 4+2T 2+T +2)X+T 6+T 4+3T 3+2T 2+T +2 [2, 3,0, 4, 2,3, 9, 2] − − − − − Y 2=X3+(3T 4+2T 3+T 2+T +2)X+T 6+T 5+3T 4+3T 2+T +3 [1,2, 1,3, 2, 1,6, 8] − − − − Y 2=X3+(3T 4+2T 3+3T 2+T +3)X+T 6+T 5+2T 3+3T 2+3T +1 [ 3,0, 3, 1, 4,9,2,6] − − − − Y 2=X3+(3T 4+3T 3+3T 2+T +3)X+T 6+4T 5+3T 4+3T 2+3T +1 [0, 1, 2, 4,4, 3,3, 2] − − − − − Y 2=X3+(3T 4+4T 3+3T 2+4T +3)X+T 6+2T 5+3T 4+T 3+2T 2+T +4 [ 3, 4,1, 1, 8, 3, 6, 2] − − − − − − − Y 2=X3+(3T 4+4T 3+4T 2+3)X+T 6+2T 5+3T 2+2T +4 [ 3,0,3,3,8, 1,2, 4] − − − T,T +1,T +2,T +4 Y 2=X3+(3T 4+4T 2+2T +3)X+T 6+T 5+T 4+3T 2+T +1 [2, 2, 6,2, 2,2,2,2] − − − Y 2=X3+(3T 4+T 3+2T 2+4T +2)X+T 6+3T 5+2T 4+4T 3+2T 2+3T +3 [ 2,2, 2, 6, 6,2, 2, 2] − − − − − − Y 2=X3+(3T 4+2T 3+2T 2+3T +3)X+T 6+T 4+4T 3+4T 2+4 [2,2,2,2, 2, 2, 6, 2] − − − − Y 2=X3+(3T 4+3T 3+4T 2+3)X+T 6+T 5+T 4+4T 3+T 2+4T +4 [2, 6, 2, 2,2, 2,2, 6] − − − − − Y 2=X3+(3T 4+4T 3+3T 2+T +2)X+T 6+T 4+T +3 [2, 2,2, 2,2, 6, 2,2] − − − − T,T +3,T 2+T +1 Y 2=X3+(3T 4+T 3+3T 2+T +3)X+T 6+3T 5+3T 2+3T +1 [ 1, 4, 3,7, 8,3, 7, 3] − − − − − − Y 2=X3+(3T 4+2T 3+3T 2+2)X+T 6+3T 5+4T 4+4T 3+2T 2+2 [2,2,0,4, 8,6,8, 6] − − Y 2=X3+(3T 4+3T 3+3T 2+3)X+T 6+4T 5+T 3+T 2+4 [ 3, 2, 1, 3, 4, 7,7,9] − − − − − − Y 2=X3+(3T 4+4T 3+T 2+T +2)X+T 6+2T 5+2T 2+4T +2 [ 3,0, 3, 7, 4, 1, 1, 1] − − − − − − − T,T +4,T 2+T +1 Y 2=X3+(3T 4+T 3+3T 2+3T +2)X+T 6+3T 5+4T 2+2T +2 [3,1,2,4, 5, 2, 5, 2] − − − − Y 2=X3+(3T 4+2T 3+2T 2+2T +3)X+T 6+3T 5+2T 4+T 3+2T 2+3T +1 [ 2, 2, 2,6,6, 6, 2,2] − − − − − Continued on next page

157 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+2T 3+2T 2+4T +2)X+T 6+T 5+2T 4+4T +3 [ 3, 2, 1, 5,4,1, 2,1] − − − − − T,T +2,T 2+4T +2 Y 2=X3+(3T 4+4T +2)X+T 6+3T 5+4T 4+4T 3+3T 2+T +2 [2,1,3,7, 4, 4,2,9] − − Y 2=X3+(3T 4+4T 2+2T +3)X+T 6+2T 4+T 3+3T 2+T +1 [ 4, 3, 1, 7, 6, 8, 2, 3] − − − − − − − − Y 2=X3+(3T 4+T 3+T 2+4T +2)X+T 6+3T 5+4T 4+2T 3+2T 2+T +2 [2,0,2,8,6, 8, 8, 6] − − − Y 2=X3+(3T 4+3T 3+3T 2+2)X+T 6+T 5+2T 4+T 3+3T 2+3 [0, 3, 3, 1,2, 4,2, 1] − − − − − T,T +1,(T +3)2 Y 2=X3+,3T 4+T 2+3)X+T 6+3T 5+T 3+4T 2+1 [0,4,6, 4,6, 6, 6,4] − − − Y 2=X3+(3T 4+T 2+3T +2)X+T 6+T 4+4T 2+3T +3 [1,3,3, 8, 6,3,3, 4] − − − Y 2=X3+(3T 4+T 2+3T +2)X+T 6+3T 4+2T 3+4T 2+2T +2 [ 4, 2, 2,2,4, 2, 2, 4] − − − − − − Y 2=X3+(3T 4+T 3+2T 2+4T +3)X+T 6+3T 5+2T 4+4T 3+3T 2+4T +1 [1,1,1, 4,6, 9,1, 4] − − − Y 2=X3+(3T 4+T 3+3T 2+2)X+T 6+2T 5+T 4+T 2+2T +3 [ 2, 2, 2, 2, 10, 6, 2, 2] − − − − − − − − Y 2=X3+(3T 4+T 3+3T 2+2)X+T 6+3T 5+2T 2+2 [3,3, 7,8, 10, 1, 7,8] − − − − Y 2=X3+(3T 4+2T 3+4T 2+T +3)X+T 6+T 5+T 4+4T 3+1 [3,1,3, 4, 6,3,3, 8] − − − Y 2=X3+(3T 4+2T 3+4T 2+T +3)X+T 6+T 5+3T 4+T 2+3T +1 [ 2, 4, 2, 4,4, 2, 2,2] − − − − − − Y 2=X3+(3T 4+2T 3+4T 2+4T +2)X+T 6+3T 5+4T 3+T 2+T +2 [4,0, 6,4,6, 6,6, 4] − − − T,T 3+4T 2+4 Y 2=X3+(3T 4+4T +3)X+T 6+2T 3+T 2+3T +4 [ 1, 4, 2,0,8,2, 7, 2] − − − − − T,T 3+T +4 Y 2=X3+(3T 4+4T 3+3)X+T 6+2T 5+4T 4+3T 3+4 [ 1, 2, 4,0,2,8, 7,1] − − − − T,T 3+2T 2+T +4 Y 2=X3+(3T 4+3T 2+3T +2)X+T 6+4T 4+4T 3+2T 2+2T +3 [0,2, 4, 3,3, 2,2, 8] − − − − Y 2=X3+(3T 4+T 3+2T 2+3T +3)X+T 6+3T 5+2T 4+3T 2+T +4 [ 2,0, 4, 1, 3,4,8,2] − − − − Y 2=X3+(3T 4+3T 3+T 2+3)X+T 6+4T 5+3T 2+4T +1 [3, 3,3,0, 1,8, 4, 4] − − − − Y 2=X3+(3T 4+3T 3+2T 2+2T +3)X+T 6+4T 5+2T 4+3T 3+2T 2+2T +1 [1, 3, 1, 4, 3, 8,2,8] − − − − − Y 2=X3+(3T 4+4T 3+2T 2+3T +3)X+T 6+2T 5+T 3+3T 2+T +4 [ 3, 3, 1,0,9, 4, 6, 8] − − − − − − Y 2=X3+(3T 4+4T 3+4T 2+3T +2)X+T 6+2T 5+2T 4+3T 2+2T +2 [ 1,1,3,2, 1, 2, 4, 4] − − − − − T,T +1,T 2+2T +4 Y 2=X3+(3T 4+T 3+2T 2+3)X+T 6+3T 5+3T 3+T 2+1 [ 3, 1, 2, 4, 3, 8,1, 7] − − − − − − − Y 2=X3+(3T 4+2T 3+2T 2+3T +3)X+T 6+T 5+3T 2+T +4 [ 1, 3, 4, 8,7,2,3,3] − − − − Y 2=X3+(3T 4+3T 3+4T 2+3T +2)X+T 6+4T 5+2T 2+3T +3 [ 3, 3,0, 4, 7,2, 1, 1] − − − − − − Y 2=X3+(3T 4+4T 3+2T 2+2)X+T 6+T 5+T 4+2T 3+2T 2+3 [2,0,2, 8,4,2, 6,6] − − T,T +2,T +3,T +4 Y 2=X3+(3T 4+T 2+T +3)X+T 6+2T 5+4T 4+3T 2+2T +4 [2, 6, 2, 2,2, 2,2, 6] − − − − − Y 2=X3+(3T 4+T 3+T 2+3)X+T 6+2T 5+4T 4+2T 3+T 2+3T +1 [2, 2, 6, 2, 6,2, 2,2] − − − − − Y 2=X3+(3T 4+2T 3+3T 2+2T +2)X+T 6+T 5+2T 4+3T 3+T 2+2T +3 [ 2, 2,2,2, 2, 6, 6,2] − − − − − Y 2=X3+(3T 4+3T 3+2T 2+3T +2)X+T 6+4T 4+2T +2 [2,2, 2,2,2,2, 2, 2] − − − Y 2=X3+(3T 4+4T 3+3T 2+4T +3)X+T 6+4T 4+2T 3+4T 2+1 [2,2,2, 6, 2, 2,2, 2] − − − − T,T +3,T 2+3T +3 Y 2=X3+(3T 4+3T 3+T 2+2T +2)X+T 6+4T 5+T 4+T 3+4T 2+4T +3 [2,2,2, 6,2, 6,2,6] − − Y 2=X3+(3T 4+3T 3+4T 2+3T +3)X+T 6+4T 5+3T 4+T 3+2T 2+T +4 [ 2, 1, 3, 2, 2,1,1, 5] − − − − − − Y 2=X3+(3T 4+3T 3+4T 2+4T +2)X+T 6+4T 5+3T 4+T +2 [ 1, 2, 3,1,1, 2, 2,4] − − − − − T,T +2,T 2+2 Y 2=X3+(3T 4+T 2+T +3)X+T 6+3T 2+2T +4 [ 3, 1, 4,3, 7, 3,3,2] − − − − − Y 2=X3+(3T 4+T 3+2T 2+2T +3)X+T 6+4T 4+T 3+2T 2+T +1 [0,2,2,6,8, 6, 6,2] − − Y 2=X3+(3T 4+2T 3+3T 2+4T +2)X+T 6+T 5+T 3+2T 2+3T +3 [ 1, 3, 2, 7,7,9,1, 8] − − − − − Y 2=X3+(3T 4+3T 3+2T 2+4T +3)X+T 6+4T 5+2T 2+4T +1 [ 3, 3,0, 1, 1, 1, 1,2] − − − − − − T,T +4,T 2+4T +1 Y 2=X3+(3T 4+4T 2+4T +2)X+T 6+T 5+4T 4+T 2+4T +3 [0, 4, 2, 8, 2,6, 2, 2] − − − − − − Y 2=X3+(3T 4+T 3+T +3)X+T 6+2T 5+3T 3+2T +1 [ 2,2,2, 2, 2, 6,6,2] − − − − Y 2=X3+(3T 4+T 3+T 2+2)X+T 6+3T 5+T 4+4T 2+3 [0,2,4, 2, 8,6,4, 2] − − − Continued on next page

158 C.5. Table for non-primes of degree 4 over F5

Conductor Curve Trace Y 2=X3+(3T 4+T 3+2T 2+3T +3)X+T 6+3T 5+T 3+T 2+2T +4 [ 2,0, 4, 8,4, 2, 8, 2] − − − − − − Y 2=X3+(3T 4+2T 3+T 2+4T +3)X+T 6+4T 4+T 3+2T 2+2T +1 [2,2, 2,6, 6,2, 6, 6] − − − − Y 2=X3+(3T 4+2T 3+2T 2+3T +2)X+T 6+T 5+T 4+T 3+2T +2 [4,2,0, 2,4, 2, 8,6] − − − Y 2=X3+(3T 4+2T 3+3T 2+3T +2)X+T 6+T 5+4T 3+4T 2+2T +2 [ 4,0, 2,4, 2, 2, 2, 2] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+T +3)X+T 6+3T 5+4T 4+4T 3+2T +4 [ 2, 4,0,4, 8, 2,4,6] − − − − Y 2=X3+(3T 4+4T 3+T 2+2T +3)X+T 6+2T 5+T 4+3T 3+3T +4 [2, 2,2, 6,6,2, 2,2] − − − T,T +1,T 2+3T +4 Y 2=X3+(3T 4+2T 2+T +3)X+T 6+T 4+3T 3+3T +1 [ 1, 3, 2, 3, 4, 8,2,7] − − − − − − Y 2=X3+(3T 4+3T 2+T +2)X+T 6+4T 4+4T 3+2T 2+2T +2 [0, 2, 2,4, 8,2, 8,8] − − − − Y 2=X3+(3T 4+2T 3+T 2+2T +2)X+T 6+4T 5+2T 4+2T 3+4T 2+2T +3 [3,3,0, 7, 4,2,2, 1] − − − Y 2=X3+(3T 4+3T 3+2T 2+2T +3)X+T 6+4T 5+2T 4+4T +4 [ 3, 1, 4,7, 8,2, 2, 7] − − − − − − T,T +1,(T +2)2 Y 2=X3+,3T 4+T 2+3)X+T 6+4T 5+3T 4+3T 2+4 [0,4, 4,6,6,4, 6, 6] − − − Y 2=X3+(3T 4+2T 2+4T +2)X+T 6+2T 5+2T 4+2T 3+T 2+T +2 [ 3, 3,8, 7, 10, 4,2,2] − − − − − Y 2=X3+(3T 4+2T 2+4T +2)X+T 6+T 5+T 4+3T 3+4T 2+4T +3 [2,2, 2, 2, 10,6,2,2] − − − Y 2=X3+(3T 4+T 3+T 2+3T +2)X+T 6+4T 5+3T 4+T 3+4T 2+2T +2 [ 4,0,4, 6,6, 4, 6, 6] − − − − − Y 2=X3+(3T 4+T 3+4T 2+2T +3)X+T 6+4T 3+T 2+T +1 [3,1, 4,3, 6, 8,6, 6] − − − − Y 2=X3+(3T 4+T 3+4T 2+2T +3)X+T 6+3T 5+4T 4+2T 3+2T 2+4T +4 [ 2, 4, 4, 2,4, 8,6, 6] − − − − − − Y 2=X3+(3T 4+2T 3+4T 2+2)X+T 6+3T 5+3T 3+T 2+2 [4,2,2, 2,4, 2, 6,6] − − − Y 2=X3+(3T 4+2T 3+4T 2+2)X+T 6+T 5+3T 4+2T 2+2 [ 1, 3, 8,3, 6,8, 6,6] − − − − − Y 2=X3+(3T 4+4T 3+2T 2+T +3)X+T 6+4T 5+3T 4+4T 3+2T 2+3T +1 [1,1, 4,1,6, 4,6,6] − − T,T 3+T 2+3T +4 Y 2=X3+(3T 4+2T 3+4T 2+3T +3)X+T 6+4T 5+2T 4+4T +1 [0,0,3,0, 1, 1,2, 7] − − − T,T +3,T 2+3 Y 2=X3+(3T 4+T 3+T 2+3)X+T 6+3T 5+2T 4+4 [ 3, 1, 4,3, 7, 6, 8, 2] − − − − − − − Y 2=X3+(3T 4+T 3+2T 2+3T +2)X+T 6+T 5+4T 4+2T 3+2T +2 [1,3,2, 7,7, 4, 4,2] − − − Y 2=X3+(3T 4+2T 3+2T 2+T +3)X+T 6+T 5+4T 4+T 2+2T +4 [0,2,2,6,8,6, 8, 8] − − Y 2=X3+(3T 4+4T 3+2T 2+3T +3)X+T 6+4T 5+2T 4+4T +1 [ 3, 3,0, 1, 1,2, 4,2] − − − − − T,T 3+4T 2+3T +4 Y 2=X3+(3T 4+T 2+3T +3)X+T 6+4T 5+3T 4+4T +1 [3,3, 3,0,8, 1, 4, 4] − − − − Y 2=X3+(3T 4+2T 3+T 2+T +2)X+T 6+T 5+4T 4+T 2+T +3 [1, 3, 1, 2, 2, 1, 4,2] − − − − − − Y 2=X3+(3T 4+2T 3+2T 2+2)X+T 6+4T 5+4T 4+3T 3+3T 2+2 [0,4, 2,3, 2,3,2, 4] − − − Y 2=X3+(3T 4+2T 3+2T 2+3T +3)X+T 6+2T 5+2T 4+3T 3+2T 2+4T +1 [1, 1, 3, 4, 8, 3,2, 2] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+T +3)X+T 6+4T 5+2T 4+3T 2+2T +4 [ 2, 4,0, 1,4, 3,8,4] − − − − Y 2=X3+(3T 4+3T 3+2T 2+4T +3)X+T 6+4T 5+2T 4+4T 3+3T +4 [ 3, 1, 3,0, 4,9, 6, 6] − − − − − − T,T +3,T 2+2T +3 Y 2=X3+(3T 4+2T 2+2T +2)X+T 6+T 5+3T 4+2T 3+2T +3 [0,3,3,2, 1, 4, 7, 1] − − − − Y 2=X3+(3T 4+T 3+2)X+T 6+3T 5+4T 4+2T 3+2T 2+4T +3 [ 2, 1, 3, 4,7, 4, 3,1] − − − − − − Y 2=X3+(3T 4+T 3+4T 2+4T +2)X+T 6+3T 5+T 4+4T 3+4T 2+T +2 [ 2,0, 2,6,8, 8,4, 6] − − − − Y 2=X3+(3T 4+2T 3+4T 2+3)X+T 6+T 5+3T 4+T 3+2T 2+1 [ 4, 3, 1, 6, 7, 8,7,3] − − − − − − T,(T +1)2,T +4 Y 2=X3+(3T 4+T 2+T +2)X+T 6+2T 5+3T 3+2T 2+T +3 [4,2, 6,6, 2, 4, 8, 2] − − − − − Y 2=X3+(3T 4+T 2+T +2)X+T 6+3T 4+3T 3+3T +2 [ 1, 3, 6,6,3, 4, 8,3] − − − − − Y 2=X3+(3T 4+2T 3+3T 2+2T +3)X+T 6+3T 5+3T 4+2T 3+3T 2+3T +1 [1,1,6,6, 9, 4, 4,1] − − − Y 2=X3+(3T 4+3T 3+3T +3)X+T 6+T 4+4T 3+T 2+1 [2,2,2,2, 6, 2,6, 2] − − − Y 2=X3+(3T 4+3T 3+3T +3)X+T 6+T 5+2T 4+4T 3+2T 2+T +1 [ 3, 3,2,2, 1,8, 4, 7] − − − − − Y 2=X3+(3T 4+3T 3+4T 2+T +2)X+T 6+3T 5+3T 4+3T 3+T 2+T +2 [0,4, 6, 6, 6,4, 4, 6] − − − − − Y 2=X3+(3T 4+4T 3+T 2+2T +2)X+T 6+3T 5+3T 4+4T 3+4T 2+4T +3 [ 4,0, 6, 6, 6, 4,4,6] − − − − − Continued on next page

159 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+4T 3+4T 2+2)X+T 6+2T 5+4T 4+T 3+4T +2 [ 2, 4,6, 6, 2,2, 2, 2] − − − − − − Y 2=X3+(3T 4+4T 3+4T 2+2)X+T 6+4T 5+4T 3+4T 2+3 [3,1,6, 6,3, 8,8,3] − − T,T 3+2T 2+4T +4 Y 2=X3+(3T 4+3T 3+4T 2+2T +3)X+T 6+4T 5+2T 2+4T +1 [0,3,0,0, 1, 1,2,2] − − T,T +2,T 2+T +2 Y 2=X3+(3T 4+T 3+T 2+2T +2)X+T 6+3T 5+4T 2+2T +3 [1,2,3,1,1, 2, 2,3] − − Y 2=X3+(3T 4+3T 3+4T 2+2T +2)X+T 6+4T 4+2T 3+2T 2+2T +3 [ 2, 2, 2,2, 6, 6, 2, 6] − − − − − − − Y 2=X3+(3T 4+3T 3+4T 2+3T +3)X+T 6+4T 5+3T 4+4T 3+2T 2+T +4 [ 2, 1, 3, 2, 2,1, 5,3] − − − − − − T 2,T +2,T +3 Y 2=X3+(3T 4)X+T 6+2T 4 [1,1,6, 9,1, 4, 4,6] − − − Y 2=X3+(3T 4+T 2)X+T 6+2T 2 [3,3, 10, 1, 7,8, 4,2] − − − − Y 2=X3+(3T 4+T 2)X+T 6+T 5+T 4+2T 3 [ 2, 2, 10, 6, 2, 2,6,2] − − − − − − Y 2=X3+(3T 4+T 3+3T 2)X+T 6+4T 3 [0,4,6, 6,6, 4, 4, 6] − − − − Y 2=X3+(3T 4+T 3+3T 2+2T )X+T 6+3T 5+T 4+2T 3+3T 2+2T [1,3, 6,3,3, 8, 8, 6] − − − − Y 2=X3+(3T 4+T 3+3T 2+2T )X+T 6+3T 5+3T 4+4T 3+3T 2 [ 4, 2,4, 2, 2,2, 8, 6] − − − − − − Y 2=X3+(3T 4+4T 3+3T 2)X+T 6+T 3 [4,0,6, 6, 6,4,4, 6] − − − Y 2=X3+(3T 4+4T 3+3T 2+3T )X+T 6+2T 5+T 4+3T 3+3T 2+3T [3,1, 6,3,3, 4,8,6] − − Y 2=X3+(3T 4+4T 3+3T 2+3T )X+T 6+2T 5+3T 4+T 3+3T 2 [ 2, 4,4, 2, 2, 4, 2,6] − − − − − − T 2,T 2+T +1 Y 2=X3+(3T 4+3T 2)X+T 6+4T 5+4T 4+T 3 [2,0,0,2,4, 4, 2, 8] − − − Y 2=X3+(3T 4+T 3+3T )X+T 6+3T 5+4T 4+T 3+2T [ 2, 2, 1, 1,4, 4,7,2] − − − − − Y 2=X3+(3T 4+2T 3+3T 2)X+T 6+4T 5+T 4+4T 3 [ 2, 2,2, 4, 8,8, 2, 4] − − − − − − Y 2=X3+(3T 4+3T 3)X+T 6+4T 5+4T 4 [0, 2, 1, 3,4, 2,1,4] − − − − Y 2=X3+(3T 4+3T 3+4T 2)X+T 6+4T 5+T 3 [ 2,0, 4,0,2,2, 6,2] − − − T 2,(T +1)2 Y 2=X3+,3T 4+2T 2)X+T 6+3T 5+3T 3 [ 3, 3,0, 4,4, 7,2,8] − − − − Y 2=X3+(3T 4+2T 2)X+T 6+4T 5+2T 3 [2,2,0,6, 6, 2,2, 2] − − − Y 2=X3+(3T 4+T 3+3T 2)X+T 6+4T 4+2T 3+2T 2 [4,1, 1, 4, 1,1,6, 4] − − − − Y 2=X3+(3T 4+T 3+3T 2)X+T 6+T 5+4T 4+4T 3 [ 1,1,4, 4,4,1,6, 4] − − − Y 2=X3+(3T 4+2T 3+T )X+T 6+3T 5+2T 3+T 2+2T [0, 3, 3,8,1, 7,2, 4] − − − − Y 2=X3+(3T 4+2T 3+T )X+T 6+2T 5+3T 3+4T 2 [0,2,2, 2, 4, 2,2,6] − − − Y 2=X3+(3T 4+3T 3)X+T 6+T 5 [3, 1,2,4,8,1, 4, 6] − − − Y 2=X3+(3T 4+3T 3)X+T 6+2T 5+T 4 [ 2, 1, 3, 6,3,1, 4,4] − − − − − Y 2=X3+(3T 4+3T 3)X+T 6+4T 5+3T 4 [ 2,4,2, 1, 7,1, 9, 1] − − − − − Y 2=X3+(3T 4+4T 3+4T 2+3T )X+T 6+T [2, 1,3, 6, 3,1, 4,4] − − − − Y 2=X3+(3T 4+4T 3+4T 2+3T )X+T 6+4T 5+T 4+4T 3+T 2 [ 3, 1, 2,4, 8,1, 4, 6] − − − − − − Y 2=X3+(3T 4+4T 3+4T 2+3T )X+T 6+2T 5+3T 4+2T 3+3T 2+3T [2,4, 2, 1,7,1, 9, 1] − − − − T 2,(T +4)2 Y 2=X3+,3T 4+2T 2)X+T 6+2T 5+2T 3 [0, 3, 3, 4,4, 2, 8, 2] − − − − − − Y 2=X3+(3T 4+2T 2)X+T 6+T 5+3T 3 [0,2,2,6, 6,8,2, 2] − − Y 2=X3+(3T 4+T 3+4T 2+2T )X+T 6+4T [3, 1,2, 6, 3,2, 7, 2] − − − − − Y 2=X3+(3T 4+T 3+4T 2+2T )X+T 6+T 5+T 4+T 3+T 2 [ 2, 1, 3,4, 8,2, 2, 2] − − − − − − Y 2=X3+(3T 4+T 3+4T 2+2T )X+T 6+3T 5+3T 4+3T 3+3T 2+2T [ 2,4,2, 1,7, 3, 2,3] − − − − Y 2=X3+(3T 4+2T 3)X+T 6+3T 5+T 4 [ 3, 1, 2, 6,3, 2,7,2] − − − − − Y 2=X3+(3T 4+2T 3)X+T 6+T 5+3T 4 [2,4, 2, 1, 7,3,2, 3] − − − − Y 2=X3+(3T 4+2T 3)X+T 6+4T 5 [2, 1,3,4,8, 2,2,2] − − Y 2=X3+(3T 4+3T 3+4T )X+T 6+2T 5+3T 3+T 2+3T [ 3, 3,0,8,1, 2,1, 2] − − − − Continued on next page

160 C.5. Table for non-primes of degree 4 over F5

Conductor Curve Trace Y 2=X3+(3T 4+3T 3+4T )X+T 6+3T 5+2T 3+4T 2 [2,2,0, 2, 4, 2, 4,8] − − − − Y 2=X3+(3T 4+4T 3+3T 2)X+T 6+4T 4+3T 3+2T 2 [ 1,1,4, 4, 1, 6, 1, 6] − − − − − − Y 2=X3+(3T 4+4T 3+3T 2)X+T 6+4T 5+4T 4+T 3 [4,1, 1, 4,4, 6,4, 6] − − − − T 2,T 2+4T +1 Y 2=X3+(3T 4+3T 2)X+T 6+T 5+4T 4+4T 3 [2,0,0,2,4, 4,6,8] − Y 2=X3+(3T 4+2T 3)X+T 6+T 5+4T 4 [ 3, 1, 2,0,4, 2,3, 8] − − − − − Y 2=X3+(3T 4+2T 3+4T 2)X+T 6+T 5+4T 3 [0, 4,0, 2,2,2,2, 6] − − − Y 2=X3+(3T 4+3T 3+3T 2)X+T 6+T 5+T 4+T 3 [ 4,2, 2, 2, 8,8, 10,4] − − − − − Y 2=X3+(3T 4+4T 3+2T )X+T 6+2T 5+4T 4+4T 3+3T [ 1, 1, 2, 2,4, 4, 1,4] − − − − − − T 2,T 2+2 Y 2=X3+(3T 4)X+T 6+4T 4 [ 4,1,1, 4, 9, 4,6,1] − − − − Y 2=X3+(3T 4+2T 2)X+T 6+3T 2 [0, 3, 3,0, 1,8,2, 7] − − − − Y 2=X3+(3T 4+2T 2)X+T 6+3T 4 [0,2,2,0, 6, 2,2, 2] − − − Y 2=X3+(3T 4+2T 3)X+T 6+T 5+2T 4 [2, 1,3,0,3, 2, 8, 3] − − − − Y 2=X3+(3T 4+3T 3)X+T 6+4T 5+2T 4 [0,3, 1,2,3, 4,6,9] − − T 2,T 2+T +2 Y 2=X3+(3T 4+T 3+T )X+T 6+3T 5+4T 4+4T 2 [ 3,1, 4,0,9,4, 3, 1] − − − − Y 2=X3+(3T 4+T 3+2T 2)X+T 6+4T 4+2T 3 [3,1, 2,2,3, 8, 3,9] − − − Y 2=X3+(3T 4+T 3+3T 2)X+T 6+2T 5+2T 4+4T 3 [0,4,1, 1,6,4, 9, 9] − − − Y 2=X3+(3T 4+2T 3+2T 2)X+T 6+T 5+2T 4+2T 3 [ 4,0, 1, 1, 6, 4, 1,1] − − − − − − Y 2=X3+(3T 4+3T 3)X+T 6 [0, 2, 1, 3, 6, 2, 9, 1] − − − − − − − Y 2=X3+(3T 4+3T 3)X+T 6+4T 5+4T 4+3T 3 [0, 2,4,2, 6, 2,6, 6] − − − − Y 2=X3+(3T 4+3T 3+3T 2)X+T 6+4T 5+4T 3 [1, 3, 2, 2, 3,8,5,7] − − − − Y 2=X3+(3T 4+4T 3)X+T 6+4T 5 [3, 3,0,4,3,4, 3, 3] − − − T 2,T +3,T +4 Y 2=X3+(3T 4)X+T 6+2T 5+3T 4 [1,1, 4,1, 4,1,6, 4] − − − Y 2=X3+(3T 4+T 3+3T 2)X+T 6+2T 5+3T 4+T 3 [0,4, 4, 6, 4,6,6,4] − − − Y 2=X3+(3T 4+T 3+4T 2)X+T 6+2T 4+T 3+2T 2 [ 3, 3, 4, 7,8, 7, 10, 4] − − − − − − − Y 2=X3+(3T 4+T 3+4T 2)X+T 6+2T 5+2T 4+4T 3 [2,2,6, 2, 2, 2, 10,6] − − − − Y 2=X3+(3T 4+2T 3+2T )X+T 6+T 5+4T 4+T 2 [ 2, 4, 2, 2, 4, 2,4, 8] − − − − − − − Y 2=X3+(3T 4+2T 3+2T )X+T 6+3T 5+4T 3+2T 2+3T [3,1,8,3, 4,3, 6, 8] − − − Y 2=X3+(3T 4+2T 3+2T 2)X+T 6+2T 5+3T 4+2T 3 [ 4,0,4,6,4, 6,6, 4] − − − Y 2=X3+(3T 4+3T 3+4T 2+2T )X+T 6+T 5+3T 3+3T 2 [4,2, 8, 2,2, 2,4, 2] − − − − Y 2=X3+(3T 4+3T 3+4T 2+2T )X+T 6+4T 5+3T 4+T 3+4T 2+4T [ 1, 3, 8,3, 8,3, 6,8] − − − − − T 2,T +1,T +2 Y 2=X3+(3T 4)X+T 6+3T 5+3T 4 [1,1, 4,1, 9, 4,6,6] − − − Y 2=X3+(3T 4+2T 3+4T 2+3T )X+T 6+4T 5+2T 3+3T 2 [2,4, 8, 2, 2, 4, 6,6] − − − − − Y 2=X3+(3T 4+2T 3+4T 2+3T )X+T 6+T 5+3T 4+4T 3+4T 2+T [ 3, 1, 8,3,3, 4, 6,6] − − − − − Y 2=X3+(3T 4+3T 3+3T )X+T 6+2T 5+T 3+2T 2+2T [1,3,8,3,3, 8,6, 6] − − Y 2=X3+(3T 4+3T 3+3T )X+T 6+4T 5+4T 4+T 2 [ 4, 2, 2, 2, 2,2,6, 6] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2)X+T 6+3T 5+3T 4+3T 3 [0, 4,4,6, 6, 4, 6, 6] − − − − − Y 2=X3+(3T 4+4T 3+3T 2)X+T 6+3T 5+3T 4+4T 3 [4,0, 4, 6, 6,4, 6, 6] − − − − − Y 2=X3+(3T 4+4T 3+4T 2)X+T 6+2T 4+4T 3+2T 2 [ 3, 3, 4, 7, 1,8,2,2] − − − − − Y 2=X3+(3T 4+4T 3+4T 2)X+T 6+T 5+T 4 [2,2,6, 2, 6, 2,2,2] − − − T 2,T 2+4T +2 Y 2=X3+(3T 4+T 3)X+T 6+T 5 [4,0, 3,3,3,4, 2,2] − − Y 2=X3+(3T 4+2T 3)X+T 6 [ 3, 1, 2,0, 6, 2, 2,7] − − − − − − Continued on next page

161 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+2T 3)X+T 6+T 5+4T 4+2T 3 [2,4, 2,0, 6, 2, 2,2] − − − − Y 2=X3+(3T 4+2T 3+3T 2)X+T 6+T 5+T 3 [ 2, 2, 3,1, 3,8,4, 2] − − − − − Y 2=X3+(3T 4+3T 3+2T 2)X+T 6+4T 5+2T 4+3T 3 [ 1, 1,0, 4, 6, 4,4,1] − − − − − Y 2=X3+(3T 4+4T 3+4T )X+T 6+2T 5+4T 4+4T 2 [0, 4,1, 3,9,4, 2, 2] − − − − Y 2=X3+(3T 4+4T 3+2T 2)X+T 6+4T 4+3T 3 [2, 2,1,3,3, 8, 8,2] − − − Y 2=X3+(3T 4+4T 3+3T 2)X+T 6+3T 5+2T 4+T 3 [ 1,1,4,0,6,4,4, 1] − − T 2,T 2+3 Y 2=X3+(3T 4)X+T 6+T 4 [1, 4, 4,1, 9, 4,1,6] − − − − Y 2=X3+(3T 4+3T 2)X+T 6+3T 2 [ 3,0,0, 3, 1, 4, 7,2] − − − − − Y 2=X3+(3T 4+3T 2)X+T 6+2T 4 [2,0,0,2, 6,6, 2,2] − − Y 2=X3+(3T 4+T 3)X+T 6+3T 5+3T 4 [ 1,0,2,3,3, 2, 3,6] − − − Y 2=X3+(3T 4+4T 3)X+T 6+2T 5+3T 4 [3,2,0, 1,3,2,9, 8] − − T 2,T +2,T +4 Y 2=X3+(3T 4)X+T 6+T 5+2T 4 [1,1,1, 4, 4,6, 4, 9] − − − − Y 2=X3+(3T 4+T 3+4T )X+T 6+3T 5+T 4+T 2 [ 4, 2, 2, 2, 8,4,2, 2] − − − − − − Y 2=X3+(3T 4+T 3+4T )X+T 6+4T 5+3T 3+2T 2+4T [1,3,3,8, 8, 6, 8,3] − − − Y 2=X3+(3T 4+T 3+3T 2)X+T 6+T 5+2T 4+4T 3 [0, 4,6,4, 4,6, 4, 6] − − − − Y 2=X3+(3T 4+3T 3+T 2)X+T 6+3T 4+2T 3+2T 2 [ 3, 3, 7, 4, 4, 10,8, 1] − − − − − − − Y 2=X3+(3T 4+3T 3+T 2)X+T 6+T 5+3T 4+3T 3 [2,2, 2,6,6, 10, 2, 6] − − − − Y 2=X3+(3T 4+3T 3+2T 2)X+T 6+T 5+2T 4+2T 3 [4,0, 6, 4,4,6,4, 6] − − − Y 2=X3+(3T 4+4T 3+T 2+4T )X+T 6+2T 5+2T 4+2T 3+4T 2+2T [ 3, 1,3, 8,8, 6, 4,3] − − − − − Y 2=X3+(3T 4+4T 3+T 2+4T )X+T 6+3T 5+T 3+3T 2 [2,4, 2, 8, 2,4, 4, 2] − − − − − T 2,T 2+2T +3 Y 2=X3+(3T 4+T 3)X+T 6 [ 1,0, 3, 2, 2, 6, 2,7] − − − − − − Y 2=X3+(3T 4+T 3)X+T 6+3T 5+T 4+4T 3 [4,0,2, 2, 2, 6, 2,2] − − − − Y 2=X3+(3T 4+T 3+2T 2)X+T 6+3T 5+2T 3 [ 2,1, 2, 3,8, 3, 8, 9] − − − − − − Y 2=X3+(3T 4+2T 3+3T )X+T 6+T 5+T 4+4T 2 [ 4, 3,0,1,4,9, 8, 5] − − − − Y 2=X3+(3T 4+2T 3+2T 2)X+T 6+4T 5+3T 4+2T 3 [1,0, 1,4,4,6,4, 1] − − Y 2=X3+(3T 4+2T 3+3T 2)X+T 6+T 4+T 3 [ 2,3,2,1, 8,3,4, 7] − − − Y 2=X3+(3T 4+3T 3)X+T 6+3T 5 [0,3,4, 3,4,3, 4,1] − − Y 2=X3+(3T 4+4T 3+3T 2)X+T 6+2T 5+3T 4+T 3 [ 1, 4, 1,0, 4, 6,4,9] − − − − − T 2,T 2+3T +3 Y 2=X3+(3T 4+T 3+3T 2)X+T 6+3T 5+3T 4+4T 3 [0, 1, 4, 1, 4, 6, 9,1] − − − − − − Y 2=X3+(3T 4+2T 3)X+T 6+2T 5 [ 3,4,3,0,4,3, 7, 3] − − − Y 2=X3+(3T 4+3T 3+2T )X+T 6+4T 5+T 4+4T 2 [1,0, 3, 4,4,9,1, 1] − − − Y 2=X3+(3T 4+3T 3+2T 2)X+T 6+T 5+3T 4+3T 3 [4, 1,0,1,4,6, 1, 9] − − − Y 2=X3+(3T 4+3T 3+3T 2)X+T 6+T 4+4T 3 [1,2,3, 2, 8,3,5,9] − − Y 2=X3+(3T 4+4T 3)X+T 6 [ 2, 3,0, 1, 2, 6,7, 1] − − − − − − Y 2=X3+(3T 4+4T 3)X+T 6+2T 5+T 4+T 3 [ 2,2,0,4, 2, 6,2, 6] − − − − Y 2=X3+(3T 4+4T 3+2T 2)X+T 6+2T 5+3T 3 [ 3, 2,1, 2,8, 3, 3,7] − − − − − T 2,T +1,T +3 Y 2=X3+(3T 4)X+T 6+4T 5+2T 4 [1,1,1, 4,6,6,1, 4] − − Y 2=X3+(3T 4+T 3+T 2+T )X+T 6+2T 5+4T 3+3T 2 [4,2, 2, 8,6, 6, 2,2] − − − − Y 2=X3+(3T 4+T 3+T 2+T )X+T 6+3T 5+2T 4+3T 3+4T 2+3T [ 1, 3,3, 8,6, 6,3, 8] − − − − − Y 2=X3+(3T 4+2T 3+T 2)X+T 6+3T 4+3T 3+2T 2 [ 3, 3, 7, 4,2,2, 7,8] − − − − − Y 2=X3+(3T 4+2T 3+T 2)X+T 6+3T 5+4T 4 [2,2, 2,6,2,2, 2, 2] − − − Continued on next page

162 C.5. Table for non-primes of degree 4 over F5

Conductor Curve Trace Y 2=X3+(3T 4+2T 3+2T 2)X+T 6+4T 5+2T 4+3T 3 [0,4, 6, 4, 6, 6,6, 4] − − − − − Y 2=X3+(3T 4+4T 3+T )X+T 6+T 5+2T 3+2T 2+T [3,1,3,8, 6,6,3, 4] − − Y 2=X3+(3T 4+4T 3+T )X+T 6+2T 5+T 4+T 2 [ 2, 4, 2, 2, 6,6, 2, 4] − − − − − − − Y 2=X3+(3T 4+4T 3+3T 2)X+T 6+4T 5+2T 4+T 3 [ 4,0,6,4, 6, 6, 6,4] − − − − T 2,T +1,T +4 Y 2=X3+(3T 4)X+T 6+3T 4 [1,1, 9,6,6, 4, 4,1] − − − Y 2=X3+(3T 4+4T 2)X+T 6+2T 2 [3,3, 1, 10,2, 4,8, 7] − − − − Y 2=X3+(3T 4+4T 2)X+T 6+2T 5+4T 4+T 3 [ 2, 2, 6, 10,2,6, 2, 2] − − − − − − Y 2=X3+(3T 4+2T 3+2T 2)X+T 6+2T 3 [0,4, 6,6, 6,4, 4,6] − − − Y 2=X3+(3T 4+2T 3+2T 2+T )X+T 6+T 5+2T 4+2T 3+3T 2 [ 4, 2, 2,4,6, 2,2, 2] − − − − − Y 2=X3+(3T 4+2T 3+2T 2+T )X+T 6+T 5+4T 4+T 3+3T 2+4T [1,3,3, 6,6,8, 8,3] − − Y 2=X3+(3T 4+3T 3+2T 2)X+T 6+3T 3 [4,0, 6,6, 6, 4,4, 6] − − − − Y 2=X3+(3T 4+3T 3+2T 2+4T )X+T 6+4T 5+2T 4+3T 3+3T 2 [ 2, 4, 2,4, 6, 8, 4, 2] − − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+4T )X+T 6+4T 5+4T 4+4T 3+3T 2+T [3,1,3, 6, 6, 8, 4,3] − − − − T 2,(T +3)2 Y 2=X3+,3T 4+3T 2)X+T 6+2T 5+4T 3 [2,0,2, 6,6, 4, 2,2] − − − Y 2=X3+(3T 4+3T 2)X+T 6+4T 5+T 3 [ 3,0, 3,4, 4,1,8, 8] − − − − Y 2=X3+(3T 4+T 3+2T )X+T 6+T 5+T 3+4T 2 [0,2,2, 4, 2, 6,6, 4] − − − − Y 2=X3+(3T 4+T 3+2T )X+T 6+4T 5+4T 3+T 2+T [0, 3, 3,1,8,4, 4,1] − − − Y 2=X3+(3T 4+2T 3+T 2+T )X+T 6+3T [2,3, 1, 3, 6,8,4, 7] − − − − Y 2=X3+(3T 4+2T 3+T 2+T )X+T 6+T 5+2T 4+4T 3+3T 2+4T [2, 2,4,7, 1, 7, 1, 2] − − − − − Y 2=X3+(3T 4+2T 3+T 2+T )X+T 6+2T 5+4T 4+3T 3+T 2 [ 3, 2, 1, 8,4,3, 6, 2] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2)X+T 6+T 4+4T 3+2T 2 [4, 1,1, 1, 4,4, 4, 1] − − − − − Y 2=X3+(3T 4+3T 3+2T 2)X+T 6+3T 5+T 4+3T 3 [ 1,4,1,4, 4, 1, 4,4] − − − − Y 2=X3+(3T 4+4T 3)X+T 6+T 5+4T 4 [ 2, 3, 1,3, 6, 8,4,7] − − − − − Y 2=X3+(3T 4+4T 3)X+T 6+2T 5+2T 4 [ 2,2,4, 7, 1,7, 1,2] − − − − Y 2=X3+(3T 4+4T 3)X+T 6+3T 5 [3,2, 1,8,4, 3, 6,2] − − − T 2,T 2+2T +4 Y 2=X3+(3T 4+2T 2)X+T 6+3T 5+T 4+3T 3 [0,2,2,0, 4,4, 2,2] − − Y 2=X3+(3T 4+T 3)X+T 6+3T 5+T 4 [ 1,0, 3, 2, 2,4, 2,7] − − − − − Y 2=X3+(3T 4+T 3+T 2)X+T 6+3T 5+3T 3 [ 4, 2,0,0,2,2,6, 2] − − − Y 2=X3+(3T 4+2T 3+4T )X+T 6+T 5+T 4+3T 3+4T [ 1, 2, 1, 2, 4,4, 4, 7] − − − − − − − Y 2=X3+(3T 4+4T 3+2T 2)X+T 6+3T 5+4T 4+2T 3 [2, 2, 4, 2,8, 8,2,2] − − − − T 2,T 2+3T +4 Y 2=X3+(3T 4+2T 2)X+T 6+2T 5+T 4+2T 3 [0,2,2,0, 4,4, 2, 8] − − − Y 2=X3+(3T 4+T 3+2T 2)X+T 6+2T 5+4T 4+3T 3 [ 2, 4, 2,2,8, 8,2, 4] − − − − − Y 2=X3+(3T 4+3T 3+T )X+T 6+4T 5+T 4+2T 3+T [ 2, 1, 2, 1, 4,4,2,2] − − − − − Y 2=X3+(3T 4+4T 3)X+T 6+2T 5+T 4 [ 2, 3,0, 1, 2,4, 8,4] − − − − − Y 2=X3+(3T 4+4T 3+T 2)X+T 6+2T 5+2T 3 [0,0, 2, 4,2,2, 2,2] − − − T 2,(T +2)2 Y 2=X3+,3T 4+3T 2)X+T 6+3T 5+T 3 [2,0,2, 6,6, 4, 2,2] − − − Y 2=X3+(3T 4+3T 2)X+T 6+T 5+4T 3 [ 3,0, 3,4, 4,1, 2,2] − − − − Y 2=X3+(3T 4+T 3)X+T 6+2T 5 [ 1,2,3,8,4, 7,2, 4] − − − Y 2=X3+(3T 4+T 3)X+T 6+3T 5+2T 4 [4,2, 2, 7, 1, 2, 3, 9] − − − − − − Y 2=X3+(3T 4+T 3)X+T 6+4T 5+4T 4 [ 1, 3, 2,3, 6, 2,2, 4] − − − − − − Y 2=X3+(3T 4+2T 3+2T 2)X+T 6+T 4+T 3+2T 2 [1, 1,4, 1, 4,4, 6,6] − − − − Continued on next page

163 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+2T 3+2T 2)X+T 6+2T 5+T 4+2T 3 [1,4, 1,4, 4, 1, 6,6] − − − − Y 2=X3+(3T 4+3T 3+T 2+4T )X+T 6+2T [ 1,3,2, 3, 6,2, 2, 4] − − − − − Y 2=X3+(3T 4+3T 3+T 2+4T )X+T 6+3T 5+4T 4+2T 3+T 2 [ 1, 2, 3, 8,4,7, 2, 4] − − − − − − Y 2=X3+(3T 4+3T 3+T 2+4T )X+T 6+4T 5+2T 4+T 3+3T 2+T [4, 2,2,7, 1,2,3, 9] − − − Y 2=X3+(3T 4+4T 3+3T )X+T 6+T 5+T 3+T 2+4T [ 3, 3,0,1,8, 8, 2,2] − − − − Y 2=X3+(3T 4+4T 3+3T )X+T 6+4T 5+4T 3+4T 2 [2,2,0, 4, 2,2,8,2] − − T,T +1,T 2+4T +1 Y 2=X3+(3T 4+3T 3+2T 2+T +2)X+T 6+4T 5+2T 4+T +3 [ 1, 2, 3, 5,4,3, 5, 2] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+3T +3)X+T 6+T 5+T 4+T 2+T +1 [ 2, 2, 2,6,6, 6, 2, 6] − − − − − − Y 2=X3+(3T 4+4T 3+3T 2+2T +2)X+T 6+2T 5+4T 2+3T +2 [2,1,3,4, 5,3, 2,1] − − T,T 3+T 2+1 Y 2=X3+(3T 4+T +3)X+T 6+3T 3+T 2+2T +4 [0, 2, 4, 1,8,2,2,4] − − − T,(T +2)2,T +4 Y 2=X3+(3T 4+T 2+3)X+T 6+2T 5+4T 3+4T 2+1 [4,0,6, 4, 4, 6,4, 6] − − − − Y 2=X3+(3T 4+T 2+2T +2)X+T 6+T 4+4T 2+2T +3 [3,1,3, 8, 8, 6,8,6] − − − Y 2=X3+(3T 4+T 2+2T +2)X+T 6+3T 4+3T 3+4T 2+3T +2 [ 2, 4, 2,2, 8, 6, 2,6] − − − − − − Y 2=X3+(3T 4+3T 3+4T 2+T +2)X+T 6+2T 5+T 3+T 2+4T +2 [0,4, 6,4,4, 6, 4, 6] − − − − Y 2=X3+(3T 4+3T 3+4T 2+4T +3)X+T 6+4T 5+T 4+T 3+1 [1,3,3, 4,8,6, 8, 6] − − − Y 2=X3+(3T 4+3T 3+4T 2+4T +3)X+T 6+4T 5+3T 4+T 2+2T +1 [ 4, 2, 2, 4, 2,6, 8, 6] − − − − − − − Y 2=X3+(3T 4+4T 3+2T 2+T +3)X+T 6+2T 5+2T 4+T 3+3T 2+T +1 [1,1,1, 4, 4,6, 4,6] − − − Y 2=X3+(3T 4+4T 3+3T 2+2)X+T 6+T 5+T 4+T 3+2T 2+2 [ 2, 2, 2, 2,6,2,6,2] − − − − Y 2=X3+(3T 4+4T 3+3T 2+2)X+T 6+2T 5+2T 2+2 [3,3, 7,8, 4,2, 4,2] − − − T,T +3,T 2+T +2 Y 2=X3+(3T 4+T +2)X+T 6+2T 5+4T 4+T 3+3T 2+4T +2 [3,1,2,7, 4, 7, 3,1] − − − Y 2=X3+(3T 4+4T 2+3T +3)X+T 6+2T 4+4T 3+3T 2+4T +1 [ 1, 3, 4, 7, 6,3,7,3] − − − − − Y 2=X3+(3T 4+2T 3+3T 2+2)X+T 6+4T 5+2T 4+4T 3+3T 2+3 [ 3, 3,0, 1,2, 1, 7, 1] − − − − − − Y 2=X3+(3T 4+4T 3+T 2+T +2)X+T 6+2T 5+4T 4+3T 3+2T 2+4T +2 [2,0,2,8,6,6,4, 6] − T,T 3+T +1 Y 2=X3+(3T 4+T 3+3)X+T 6+3T 5+4T 4+2T 3+4 [0, 4, 2, 1,2,8,2, 1] − − − − T,T +1,T +2,T +3 Y 2=X3+(3T 4+T 2+4T +3)X+T 6+2T 5+T 4+3T 3+4T 2+3T +1 [2, 6, 2,2,2, 6, 2, 2] − − − − − Y 2=X3+(3T 4+T 3+3T 2+T +3)X+T 6+4T 4+3T 3+4T 2+1 [2,2,2, 6, 2, 2,2, 2] − − − − Y 2=X3+(3T 4+2T 3+2T 2+2T +2)X+T 6+4T 4+3T +2 [2,2, 2, 2, 6, 2, 6,2] − − − − − Y 2=X3+(3T 4+3T 3+3T 2+3T +2)X+T 6+4T 5+2T 4+2T 3+T 2+3T +3 [ 2, 2,2, 2,2,2, 2, 6] − − − − − Y 2=X3+(3T 4+4T 3+T 2+3)X+T 6+T 5+T 4+3T 3+2T 2+4 [2, 2, 6,2, 2,2,2,2] − − − T,T +4,T 2+3T +4 Y 2=X3+(3T 4+T 3+2T 2+2)X+T 6+4T 5+T 4+3T 3+2T 2+3 [2,0,2, 8,4, 6,8, 8] − − − Y 2=X3+(3T 4+2T 3+4T 2+2T +2)X+T 6+T 5+2T 2+2T +3 [0, 3, 3, 4, 7, 1, 1,2] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+2T +3)X+T 6+4T 5+3T 2+4T +4 [ 4, 3, 1, 8,7, 3, 7, 2] − − − − − − − Y 2=X3+(3T 4+4T 3+2T 2+3)X+T 6+2T 5+2T 3+T 2+1 [ 2, 1, 3, 4, 3,9,7,2] − − − − − T,T 3+3T 2+T +1 Y 2=X3+(3T 4+3T 2+2T +2)X+T 6+4T 4+T 3+2T 2+3T +3 [ 3, 4,2,0,3, 2, 2,9] − − − − Y 2=X3+(3T 4+T 3+2T 2+2T +3)X+T 6+3T 5+4T 3+3T 2+4T +4 [0, 1, 3, 3,9, 4, 2, 1] − − − − − − Y 2=X3+(3T 4+T 3+4T 2+2T +2)X+T 6+3T 5+2T 4+3T 2+3T +2 [2,3,1, 1, 1, 2,2,1] − − − Y 2=X3+(3T 4+2T 3+T 2+3)X+T 6+T 5+3T 2+T +1 [0,3, 3,3, 1,8,8, 7] − − − Y 2=X3+(3T 4+2T 3+2T 2+3T +3)X+T 6+T 5+2T 4+2T 3+2T 2+3T +1 [ 4, 1, 3,1, 3, 8,2,3] − − − − − Y 2=X3+(3T 4+4T 3+2T 2+2T +3)X+T 6+2T 5+2T 4+3T 2+4T +4 [ 1, 4,0, 2, 3,4, 4, 9] − − − − − − T,T +4,T 2+2T +4 Y 2=X3+(3T 4+2T 2+4T +3)X+T 6+T 4+2T 3+2T +1 [ 2, 3, 1, 3, 4,9, 7,1] − − − − − − Y 2=X3+(3T 4+3T 2+4T +2)X+T 6+4T 4+T 3+2T 2+3T +2 [ 2, 2,0,4, 8, 6,6, 6] − − − − − Continued on next page

164 C.5. Table for non-primes of degree 4 over F5

Conductor Curve Trace Y 2=X3+(3T 4+2T 3+2T 2+3T +3)X+T 6+T 5+2T 4+T +4 [ 4, 1, 3,7, 8, 3,3,3] − − − − − Y 2=X3+(3T 4+3T 3+T 2+3T +2)X+T 6+T 5+2T 4+3T 3+4T 2+3T +3 [0,3,3, 7, 4, 1, 1, 1] − − − − − T,T +1,T 2+T +1 Y 2=X3+(3T 4+4T 2+T +2)X+T 6+4T 5+4T 4+T 2+T +3 [ 2, 4,0, 8, 2, 2,4, 8] − − − − − − Y 2=X3+(3T 4+T 3+T 2+3T +3)X+T 6+3T 5+T 4+2T 3+2T +4 [2, 2,2, 6,6, 6, 6,6] − − − − Y 2=X3+(3T 4+2T 3+2T 2+4T +3)X+T 6+2T 5+4T 4+T 3+3T +4 [0, 4, 2,4, 8, 2, 8, 2] − − − − − − Y 2=X3+(3T 4+3T 3+T 2+T +3)X+T 6+4T 4+4T 3+2T 2+3T +1 [ 2,2,2,6, 6,2, 2, 2] − − − − Y 2=X3+(3T 4+3T 3+2T 2+2T +2)X+T 6+4T 5+T 4+4T 3+3T +2 [0,2,4, 2,4, 2, 2, 8] − − − − Y 2=X3+(3T 4+3T 3+3T 2+2T +2)X+T 6+4T 5+T 3+4T 2+3T +2 [ 2,0, 4,4, 2,6, 8,4] − − − − Y 2=X3+(3T 4+4T 3+4T +3)X+T 6+T 5+4T 4+2T 3+4T 2+T +1 [2,2, 2, 2, 2,2,6, 6] − − − − Y 2=X3+(3T 4+4T 3+T 2+2)X+T 6+2T 5+T 4+4T 2+3 [4,2,0, 2, 8, 2, 2,4] − − − − Y 2=X3+(3T 4+4T 3+2T 2+2T +3)X+T 6+2T 5+4T 3+T 2+3T +4 [ 4,0, 2, 8,4,6,4, 2] − − − − T,T +3,T 2+2 Y 2=X3+(3T 4+T 2+4T +3)X+T 6+3T 2+3T +4 [ 4, 1, 3,3,7, 8, 6, 2] − − − − − − Y 2=X3+(3T 4+2T 3+2T 2+T +3)X+T 6+T 5+2T 2+T +1 [0, 3, 3, 1, 7, 4,2,2] − − − − − Y 2=X3+(3T 4+3T 3+3T 2+T +2)X+T 6+4T 5+4T 3+2T 2+2T +3 [ 2, 3, 1, 7, 3, 4, 4,2] − − − − − − − Y 2=X3+(3T 4+4T 3+2T 2+3T +3)X+T 6+4T 4+4T 3+2T 2+4T +1 [2,2,0,6,4, 8,6, 8] − − T,T +2,T 2+2T +3 Y 2=X3+(3T 4+2T 3+T 2+3T +2)X+T 6+T 5+T 4+4T 3+4T 2+T +3 [2,2,2, 6,2,6, 2, 6] − − − Y 2=X3+(3T 4+2T 3+4T 2+T +2)X+T 6+T 5+3T 4+4T +2 [ 3, 2, 1,1,1, 5, 2,3] − − − − − Y 2=X3+(3T 4+2T 3+4T 2+2T +3)X+T 6+T 5+3T 4+4T 3+2T 2+4T +4 [ 3, 1, 2, 2, 2,4, 5,3] − − − − − − T,(T +3)2,T +4 Y 2=X3+(3T 4+T 2+3)X+T 6+T 5+3T 4+3T 2+4 [4,0, 4,6, 4, 6, 6,4] − − − − Y 2=X3+(3T 4+2T 2+T +2)X+T 6+3T 5+2T 4+3T 3+T 2+4T +2 [ 3, 3,8, 7, 4, 7, 1,8] − − − − − − Y 2=X3+(3T 4+2T 2+T +2)X+T 6+T 5+2T 4+2T 2+T +2 [2,2, 2, 2,6, 2, 6, 2] − − − − − Y 2=X3+(3T 4+T 3+2T 2+4T +3)X+T 6+T 5+3T 4+T 3+2T 2+2T +1 [1,1, 4,1, 4,1, 9, 4] − − − − Y 2=X3+(3T 4+3T 3+4T 2+2)X+T 6+2T 5+2T 3+T 2+2 [2,4,2, 2, 8, 2, 2, 4] − − − − − Y 2=X3+(3T 4+3T 3+4T 2+2)X+T 6+4T 5+3T 4+2T 2+2 [ 3, 1, 8,3, 8,3,3, 4] − − − − − Y 2=X3+(3T 4+4T 3+T 2+2T +2)X+T 6+T 5+3T 4+4T 3+4T 2+3T +2 [0, 4,4, 6,4,6, 6, 4] − − − − Y 2=X3+(3T 4+4T 3+4T 2+3T +3)X+T 6+T 3+T 2+4T +1 [1,3, 4,3,8,3,3, 8] − − Y 2=X3+(3T 4+4T 3+4T 2+3T +3)X+T 6+2T 5+4T 4+3T 3+2T 2+T +4 [ 4, 2, 4, 2, 2, 2, 2,2] − − − − − − − T,T 3+T 2+3T +1 Y 2=X3+(3T 4+T 2+2T +3)X+T 6+T 5+3T 4+T +1 [0, 3,3,3,8, 1,8, 4] − − − Y 2=X3+(3T 4+2T 3+2T 2+T +3)X+T 6+T 5+2T 4+T 3+2T +4 [0, 3, 1, 3, 4,9, 2,6] − − − − − Y 2=X3+(3T 4+2T 3+2T 2+4T +3)X+T 6+T 5+2T 4+3T 2+3T +4 [ 1,0, 4, 2,4, 3, 4, 2] − − − − − − Y 2=X3+(3T 4+3T 3+T 2+4T +2)X+T 6+4T 5+4T 4+T 2+4T +3 [ 2, 1, 3,1, 2, 1,2, 8] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+2)X+T 6+T 5+4T 4+2T 3+3T 2+2 [3, 2,4,0, 2,3, 2, 2] − − − − Y 2=X3+(3T 4+3T 3+2T 2+2T +3)X+T 6+3T 5+2T 4+2T 3+2T 2+T +1 [ 4, 3, 1,1, 8, 3,2, 2] − − − − − − T,T +2,T 2+3 Y 2=X3+(3T 4+T 3+2T 2+2T +3)X+T 6+T 5+2T 4+T +1 [0, 3, 3, 1, 7, 1, 1,2] − − − − − − Y 2=X3+(3T 4+3T 3+2T 2+4T +3)X+T 6+4T 5+4T 4+T 2+3T +4 [2,2,0,6,4, 6, 6,2] − − Y 2=X3+(3T 4+4T 3+T 2+3)X+T 6+2T 5+2T 4+4 [ 4, 1, 3,3,7,3, 3,2] − − − − Y 2=X3+(3T 4+4T 3+2T 2+2T +2)X+T 6+4T 5+4T 4+3T 3+3T +2 [2,3,1, 7, 3,1,9, 8] − − − T,T 3+4T 2+3T +1 Y 2=X3+(3T 4+3T 3+4T 2+2T +3)X+T 6+T 5+2T 4+T +1 [0,3,0,0, 1, 1,2,2] − − T,T +2,T 2+3T +3 Y 2=X3+(3T 4+2T 2+3T +2)X+T 6+4T 5+3T 4+3T 3+3T +3 [3,3,0,2, 1, 1, 1,2] − − − Y 2=X3+(3T 4+3T 3+4T 2+3)X+T 6+4T 5+3T 4+4T 3+2T 2+1 [ 1, 3, 4, 6, 7,3, 3, 2] − − − − − − − Y 2=X3+(3T 4+4T 3+2)X+T 6+2T 5+4T 4+3T 3+2T 2+T +3 [ 3, 1, 2, 4,7, 7,9,2] − − − − − Continued on next page

165 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+4T 3+4T 2+T +2)X+T 6+2T 5+T 4+T 3+4T 2+4T +2 [ 2,0, 2,6,8,6, 6, 8] − − − − T,T +3,T 2+4T +2 Y 2=X3+(3T 4+2T 3+4T 2+2T +3)X+T 6+T 5+3T 4+T 3+2T 2+4T +4 [ 3, 1, 2, 2, 2,4, 5,1] − − − − − − Y 2=X3+(3T 4+2T 3+4T 2+3T +2)X+T 6+4T 4+3T 3+2T 2+3T +3 [ 2, 2, 2,2, 6,6,6,2] − − − − Y 2=X3+(3T 4+4T 3+T 2+3T +2)X+T 6+2T 5+4T 2+3T +3 [3,2,1,1,1, 5,4, 2] − − T,T 3+3T 2+4T +1 Y 2=X3+(3T 4+2T 3+4T 2+3T +3)X+T 6+T 5+2T 2+T +1 [0,0,3,0, 1, 1,2,8] − − T,T +1,(T +4)2 Y 2=X3+,3T 4+T 2+4T +2)X+T 6+3T 5+2T 3+2T 2+4T +3 [2,4, 6,6,4, 2,2, 2] − − − Y 2=X3+(3T 4+T 2+4T +2)X+T 6+3T 4+2T 3+2T +2 [ 3, 1, 6,6, 6,8, 8,3] − − − − − Y 2=X3+(3T 4+T 3+T 2+3T +2)X+T 6+2T 5+3T 4+T 3+4T 2+T +3 [0, 4, 6, 6,6, 4,4, 6] − − − − − Y 2=X3+(3T 4+T 3+4T 2+2)X+T 6+T 5+T 3+4T 2+3 [1,3,6, 6, 6, 8, 4,3] − − − − Y 2=X3+(3T 4+T 3+4T 2+2)X+T 6+3T 5+4T 4+4T 3+T +2 [ 4, 2,6, 6,4, 8, 4, 2] − − − − − − Y 2=X3+(3T 4+2T 3+2T +3)X+T 6+T 4+T 3+T 2+1 [2,2,2,2, 10,6, 2, 2] − − − Y 2=X3+(3T 4+2T 3+2T +3)X+T 6+4T 5+2T 4+T 3+2T 2+4T +1 [ 3, 3,2,2, 10, 4,8, 7] − − − − − Y 2=X3+(3T 4+2T 3+4T 2+4T +2)X+T 6+2T 5+3T 4+2T 3+T 2+4T +2 [4,0, 6, 6,6,4, 4,6] − − − Y 2=X3+(3T 4+3T 3+3T 2+3T +3)X+T 6+2T 5+3T 4+3T 3+3T 2+2T +1 [1,1,6,6,6, 4, 4,1] − − T,T +3,T 2+2T +4 Y 2=X3+(3T 4+2T 3+2T 2+4T +2)X+T 6+T 5+4T 2+4T +3 [2,3,1, 5,4,1,1, 2] − − Y 2=X3+(3T 4+4T 3+3T 2+T +3)X+T 6+T 5+3T 4+3T 3+2T 2+T +4 [ 2, 2, 2,6,6,2, 6, 6] − − − − − Y 2=X3+(3T 4+4T 3+3T 2+2T +2)X+T 6+2T 5+3T 4+3T +2 [ 1, 3, 2,4, 5, 2, 2,1] − − − − − − T,T +4,T 2+3T +3 Y 2=X3+(3T 4+2T +2)X+T 6+T 5+T 4+2T 3+3T 2+2T +3 [1,2,3, 4,7,1, 3,2] − − Y 2=X3+(3T 4+T 2+T +3)X+T 6+3T 4+3T 3+3T 2+2T +4 [ 3, 4, 1, 6, 7,3,7, 2] − − − − − − Y 2=X3+(3T 4+T 3+2T 2+2)X+T 6+2T 5+3T 4+3T 3+3T 2+2 [ 3,0, 3,2, 1, 1, 7,2] − − − − − Y 2=X3+(3T 4+2T 3+4T 2+2T +2)X+T 6+T 5+T 4+T 3+2T 2+2T +3 [0,2,2,6,8, 6,4, 8] − − T,T 3+3T 2+2 Y 2=X3+(3T 4+2T +3)X+T 6+T 3+T 2+T +1 [ 2, 1,0, 4,2,8, 9,1] − − − − T,(T +1)2,T +2 Y 2=X3+(3T 4+4T 2+3)X+T 6+T 5+3T 3+4T 2+4 [4,0, 4,6, 6,4, 6,6] − − − Y 2=X3+(3T 4+4T 2+4T +2)X+T 6+2T 4+T 3+4T 2+4T +3 [ 2, 4,2, 2,6, 2, 2,4] − − − − − Y 2=X3+(3T 4+4T 2+4T +2)X+T 6+4T 4+4T 2+T +2 [3,1, 8,3,6,8,3, 6] − − Y 2=X3+(3T 4+2T 3+2T 2+2)X+T 6+T 5+2T 2+3 [3,3,8, 7,2, 4, 1, 10] − − − − Y 2=X3+(3T 4+2T 3+2T 2+2)X+T 6+3T 5+4T 4+2T 3+2T 2+3 [ 2, 2, 2, 2,2,6, 6, 10] − − − − − − Y 2=X3+(3T 4+2T 3+3T 2+2T +3)X+T 6+T 5+3T 4+2T 3+3T 2+3T +4 [1,1, 4,1,6, 4, 9,6] − − − Y 2=X3+(3T 4+4T 3+T 2+2T +2)X+T 6+T 5+2T 3+T 2+2T +3 [0,4,4, 6, 6, 4, 6,6] − − − − Y 2=X3+(3T 4+4T 3+T 2+3T +3)X+T 6+2T 5+2T 4+T 2+T +4 [ 4, 2, 4, 2, 6, 8, 2,4] − − − − − − − Y 2=X3+(3T 4+4T 3+T 2+3T +3)X+T 6+2T 5+4T 4+2T 3+4 [1,3, 4,3, 6, 8,3, 6] − − − −

C.6 Table for primes of degree 4 over F5

Table C.5: Isogeny classes for primes of degree 4 over F5

Conductor Curve Trace T 4+2T 3+4T 2+3T +3 Y 2=X3+(3T 4+T 3+2T 2+4T )X+T 6+3T 5+3T 2+T +1 [0,0,0,4,0, 2,6, 6] − − T 4+4T 3+T 2+4T +3 Y 2=X3+(3T 4+2T 3+3T 2+2T )X+T 6+T 5+3T 2+2T +4 [0,4,0,0,0,6, 2, 2] − − Continued on next page

166 C.6. Table for primes of degree 4 over F5

Conductor Curve Trace T 4+2 Y 2=X3+(3T 4+2)X+T 6+3T 2 [4,0,0,0,0, 6, 6, 2] − − − T 4+T 2+2 Y 2=X3+(3T 4+3T 2)X+T 6+4T 4+4T 2+3 [0,2,0,0,2, 6,6,6] − Y 2=X3+(3T 4+2T 3+4T 2+2T )X+T 6+T 5+3T 4+T 3+3T 2+2 [0,1, 2, 2, 3,6, 3,3] − − − − Y 2=X3+(3T 4+3T 3+4T 2+3T )X+T 6+4T 5+3T 4+4T 3+3T 2+2 [0, 3, 2, 2,1,6, 3,3] − − − − T 4+2T 3+T 2+2 Y 2=X3+(3T 4+T 3+3T 2+2)X+T 6+3T 5+3T 4+T 3+3T 2+3T [4,4, 2,0, 2, 2, 2, 10] − − − − − T 4+3T 3+T 2+2 Y 2=X3+(3T 4+4T 3+3T 2+2)X+T 6+2T 5+3T 4+4T 3+3T 2+2T [4, 2,0, 2,4, 2, 2,6] − − − − T 4+4T 2+2 Y 2=X3+(3T 4+2T 2)X+T 6+T 4+4T 2+2 [0,0,2,2,0,6, 6, 2] − − Y 2=X3+(3T 4+T 3+T 2+4T )X+T 6+3T 5+2T 4+2T 3+3T 2+3 [0, 2, 3,1, 2, 3,6,4] − − − − Y 2=X3+(3T 4+4T 3+T 2+T )X+T 6+2T 5+2T 4+3T 3+3T 2+3 [0, 2,1, 3, 2, 3,6, 4] − − − − − T 4+T 3+4T 2+2 Y 2=X3+(3T 4+3T 3+2T 2+2)X+T 6+4T 5+2T 4+2T 3+3T 2+4T [4, 2, 2,4,0, 2, 2, 2] − − − − − T 4+4T 3+4T 2+2 Y 2=X3+(3T 4+2T 3+2T 2+2)X+T 6+T 5+2T 4+3T 3+3T 2+T [4,0,4, 2, 2, 2, 2,2] − − − − T 4+3T 3+T +2 Y 2=X3+(3T 4+T 3+3T 2+3T +4)X+T 6+3T 5+3T 4+T 2+2T +1 [ 2, 3,0,1, 2,3, 4, 7] − − − − − Y 2=X3+(3T 4+2T 3+4T 2+T +4)X+T 6+T 5+3T 4+3T 3+4T 2+3T +1 [ 2,1,0, 3, 2,3,4,1] − − − Y 2=X3+(3T 4+4T 3+3T )X+T 6+2T 5+4T 4+2T 3+T +2 [0,2,0,2,0,6, 2,2] − T 4+4T 3+2T +2 Y 2=X3+(3T 4+T 3+T 2+2T +4)X+T 6+3T 5+2T 4+T 3+4T 2+4T +4 [ 2,0, 2,1, 3,4,3,4] − − − Y 2=X3+(3T 4+2T 3+T )X+T 6+T 5+T 4+4T 3+3T +3 [0,0,0,2,2, 2,6, 2] − − Y 2=X3+(3T 4+3T 3+2T 2+T +4)X+T 6+4T 5+2T 4+T 2+T +4 [ 2,0, 2, 3,1, 4,3, 4] − − − − − T 4+T 3+3T +2 Y 2=X3+(3T 4+2T 3+2T 2+4T +4)X+T 6+T 5+2T 4+T 2+4T +4 [ 2,1, 3, 2,0, 4,3,4] − − − − Y 2=X3+(3T 4+3T 3+4T )X+T 6+4T 5+T 4+T 3+2T +3 [0,2,2,0,0, 2,6, 2] − − Y 2=X3+(3T 4+4T 3+T 2+3T +4)X+T 6+2T 5+2T 4+4T 3+4T 2+T +4 [ 2, 3,1, 2,0,4,3, 4] − − − − T 4+2T 3+4T +2 Y 2=X3+(3T 4+T 3+2T )X+T 6+3T 5+4T 4+3T 3+4T +2 [0,0,2,0,2,6, 2, 6] − − Y 2=X3+(3T 4+3T 3+4T 2+4T +4)X+T 6+4T 5+3T 4+2T 3+4T 2+2T +1 [ 2, 2, 3,0,1,3,4,6] − − − Y 2=X3+(3T 4+4T 3+3T 2+2T +4)X+T 6+2T 5+3T 4+T 2+3T +1 [ 2, 2,1,0, 3,3, 4,6] − − − − T 4+T 3+T 2+T +4 Y 2=X3+(3T 4+3T 3+3T 2+3T +1)X+T 6+4T 5+2T 2+3 [2,2,2,2, 4, 2,6, 2] − − − T 4+4T 3+2T 2+T +4 Y 2=X3+(3T 4+T 2+2T +3)X+T 6+3T 4+T 3+3T 2+T +2 [1,0, 3, 2, 2, 7,4,3] − − − − Y 2=X3+(3T 4+2T 3+T 2+3T +1)X+T 6+T 5+4T 4+T 3+3T 2+2 [2,0,2,0,0,2, 2,6] − Y 2=X3+(3T 4+4T 3+3T 2+3T +1)X+T 6+2T 5+3T 4+3T 3+4T 2+2T +1 [ 3,0,1, 2, 2,1, 4,3] − − − − T 4+2T 3+2T 2+T +1 Y 2=X3+(3T 4+T 3+T 2+3T +4)X+T 6+3T 5+2T 4+4T 3+4T 2+1 [ 2, 2,4,0,4,2,6, 6] − − − T 4+2T 3+3T 2+2T +4 Y 2=X3+(3T 4+4T 2+4T +3)X+T 6+2T 4+2T 3+3T 2+3T +3 [1, 3, 2,0, 2,4, 7, 3] − − − − − Y 2=X3+(3T 4+T 3+4T 2+T +1)X+T 6+3T 5+T 4+2T 3+3T 2+3 [2,2,0,0,0, 2,2,6] − Y 2=X3+(3T 4+2T 3+2T 2+T +1)X+T 6+T 5+2T 4+T 3+4T 2+T +4 [ 3,1, 2,0, 2, 4,1, 3] − − − − − T 4+3T 3+4T 2+2T +4 Y 2=X3+(3T 4+4T 3+2T 2+T +1)X+T 6+2T 5+2T 2+2 [2,2, 4,2,2,6, 2, 2] − − − T 4+T 3+3T 2+2T +1 Y 2=X3+(3T 4+3T 3+4T 2+T +4)X+T 6+4T 5+3T 4+3T 3+4T 2+4 [ 2,4,4, 2,0,6,2,6] − − T 4+3T 3+3T 2+3T +4 Y 2=X3+(3T 4+4T 2+T +3)X+T 6+2T 4+3T 3+3T 2+2T +3 [1, 2,0, 2, 3,4, 7,4] − − − − Y 2=X3+(3T 4+3T 3+2T 2+4T +1)X+T 6+4T 5+2T 4+4T 3+4T 2+4T +4 [ 3, 2,0, 2,1, 4,1, 4] − − − − − Y 2=X3+(3T 4+4T 3+4T 2+4T +1)X+T 6+2T 5+T 4+3T 3+3T 2+3 [2,0,0,0,2, 2,2, 2] − − T 4+2T 3+4T 2+3T +4 Y 2=X3+(3T 4+T 3+2T 2+4T +1)X+T 6+3T 5+2T 2+2 [2,2,2, 4,2,6, 2, 10] − − − T 4+4T 3+T 2+4T +4 Y 2=X3+(3T 4+2T 3+3T 2+2T +1)X+T 6+T 5+2T 2+3 [2, 4,2,2,2, 2,6,6] − − T 4+T 3+2T 2+4T +4 Y 2=X3+(3T 4+T 2+3T +3)X+T 6+3T 4+4T 3+3T 2+4T +2 [1, 2, 2, 3,0, 7,4,1] − − − − Y 2=X3+(3T 4+T 3+3T 2+2T +1)X+T 6+3T 5+3T 4+2T 3+4T 2+3T +1 [ 3, 2, 2,1,0,1, 4, 7] − − − − − Continued on next page

167 C. Tables

Conductor Curve Trace Y 2=X3+(3T 4+3T 3+T 2+2T +1)X+T 6+4T 5+4T 4+4T 3+3T 2+2 [2,0,0,2,0,2, 2,2] − T 4+4T 3+3T 2+3T +1 Y 2=X3+(3T 4+2T 3+4T 2+4T +4)X+T 6+T 5+3T 4+2T 3+4T 2+4 [ 2,0, 2,4,4,6,2, 2] − − − T 4+3 Y 2=X3+(3T 4+3)X+T 6+2T 2 [ 4,2,2,2,2, 10, 10,6] − − − T 4+2T 2+3 Y 2=X3+(3T 4+T 2)X+T 6+3T 4+T 2+4 [0, 2,4,4, 2, 10, 6, 2] − − − − − T 4+3T 2+3 Y 2=X3+(3T 4+4T 2)X+T 6+2T 4+T 2+1 [0,4, 2, 2,4, 6, 10,2] − − − − T 4+3T 3+2T 2+4T +1 Y 2=X3+(3T 4+4T 3+T 2+2T +4)X+T 6+2T 5+2T 4+T 3+4T 2+1 [ 2,4,0,4, 2,2,6, 2] − − − T 4+T 3+T 2+T +3 Y 2=X3+(3T 4+3T 3+3T 2+3T )X+T 6+4T 5+3T 2+3T +4 [0,0,0,0,4,6, 2,6] − T 4+3T 3+4T 2+2T +3 Y 2=X3+(3T 4+4T 3+2T 2+T )X+T 6+2T 5+3T 2+4T +1 [0,0,4,0,0, 2,6,6] −

C.7 Table for primes of degree 5 over F5

Table C.6: Isogeny classes for primes of degree 5 over F5

Conductor Curve Trace T 5+2T 4+3T 2+3T Y 2=X3+(3T 4+2T 3+4T 2+T +3X)+T 6+T 5+3T 4+3T 3+T 2+3T +1 [ 1, 2,0,1, 2, 4, 1, 2] − − − − − − T 5+T 4+2T +2 Y 2=X3+(3T 4+4T 3+4T +3X)+T 6+2T 5+4T 4+2T 3+4T 2+T +2 [1, 3,2, 2, 2, 2, 4,7] − − − − − T 5+3T 4+2T +2 Y 2=X3+(3T 4+2T 2X)+T 6+T 5+3T 4+2T 3+T 2+4T +4 [0,0, 3, 3,0, 4, 1,8] − − − − T 5+4T 4+2T +2 Y 2=X3+(3T 4+T 3+3T 2+3T +1X)+T 6+3T 5+3T 4+3T 3+T 2+4T [2, 4,0,2, 6, 6, 8,2] − − − − T 5+2T 4+T 3+2T 2+1 Y 2=X3+(3T 4+4T 3+3T 2+4T +1X)+T 6+2T 5+3T 4+T 3+T 2+2T +4 [ 3, 2, 3, 3,0, 6,2, 2] − − − − − − T 5+T 4+T 3+2T +2 Y 2=X3+(3T 4+T 3+2T 2+T +3X)+T 6+3T 5+T 3+T 2+2T [ 4, 3, 2, 2,0,2,2,1] − − − − T 5+2T 3+2T +2 Y 2=X3+(3T 4+4T 3+3T 2+3T +1X)+T 6+2T 5+3T 4+T 3+2T 2+T [2,0, 2, 2,6, 4, 2, 6] − − − − − T 5+T 4+2T 3+2T +2 Y 2=X3+(3T 4+2T 3+2T 2+4T +3X)+T 6+T 5+2T 4+3T 2+T +2 [1, 1,0, 2, 1,7, 4, 2] − − − − − T 5+3T 4+3T 3+3T 2+3T Y 2=X3+(3T 4+3T 3+3T 2+4T +3X)+T 6+4T 5+4T 4+3T 3+3T 2+2T +1 [4,2,0,2, 4, 6,2,2] − − T 5+4T 4+2T 3+2T +2 Y 2=X3+(3T 4+4T 3X)+T 6+2T 5+4T 4+3T 3+T 2+T +2 [0,3, 3,0, 1,8, 6, 3] − − − −

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